u 



PIIILADELrHIA 



.NAUTICAL COLLEGE, 

t South- West corner of Cliesnut and Eighth Sts. * 



\^ 




% In this long-established and well-known institution, the complete course of Naviga 
2 tion and Nautical Astronomy is taught in the most satisfactory manner, and in | 
J the shortest possible time. f 

f The course includes all the diflfereut sailings: viz., Plane, Traverse, Parallel, ♦ 

# Middle Latitude, Mercator's, and Great Circle Sailing; 
f All the various methods of finding the ♦ 

i LATITUDE OF THE SHIP, I 

I by altitudes of th( ~ -■. — ..* ^ -r>-i^ o^.,--- ah ^v^g^arious methods | 

I IlIBRARY qP CONGRESS. I , ,^. ,J 

# and regulating a# --"- ^ '*^» ^^^ Fixed # 

I Stars; All the V f , \[^\^ ^ ^ I 



I by Dead Reck | ^^^^ Ehl 



I the use of Quad] f UNITED STATES OF AMERICA.! i and Steering | 
I Compasses, I @^^<^^^^^^^^^^^^^<^<^^^| the method of # 

^1 Keeping a j0Ui.xic*x ^ _. , _ _ can at any time, # 

4 with confidence and certainty, point out on the Chart ^ 

THE SIEilX^'S I'OSITIOlSr .if^T SE.A-. I 



1?^ <•» # 

# # 
^1 Acting Masters' I^Iates, Ensigns, and Masters for the XJ. S. Navy, and Revenue and ^ 

# Mercantile Marine Ofiicers, prepared for examination, and qualified for appointment. # 

*$* ^ 

I . ^ I 

t ENGINEERS FOR THE U. S. NAVY. I 

% Candidates for admission or promotion in either the volunteer or regular service are ^ 
^ taught the course ordered by the Navy Department. 



I THE WEST POINT COURSE OF MATHEMATICS, 

S including Land and Topographical Surveying and Civil and Military 
% Engineering, with field practice in the use of Compass, Transit, Level, 
J Plain Table, &c., &c., taught in a course of special practical lessons adapted to the J 
% professional wants of oificers in the U. S. Army. 

I - ^r- GRADUATES ABE AWASDED A SPLENDID DIPLOMA. 

I Jf. B.— ''The United States Practical Navigator" will be sent by mail on the S 
receipt of $2.50. Ail communications must be addressed to the 
Authors, M. & G. W. EOCHE. 



^rA r'oTO^ 



^wC /r^^ K^. ^^ ^^ ^^ /^^ ^^^^ 








\ 






%^ '^ 



^^ > \ \ 



^i ^ v^ sS^ 



* . 



<v ^.% 



^ ♦ 



X. 



THE 



UNITED STATES 



PRACTICAL 



NAVIGATOR 



IN WHICH ALL THE CALCULATIONS ARE MADE BY 



ARITHMETIC 



" When the mariner has been tossed for many days, in thick weather, and on an unknown sea, 
he naturally avails himself of the first pause in the storm, the earliest glance of the sun, to 
take his latitude, and ascertain how far the elements have driven him from his true course." 

Daniel Webster, 




'•'^^^2^: 



MAETIN "ROCHE, M. D., 

PRESIDENT OP THE PHILADELPHIA NAUTICAL COLLEGE, 

AND 

aEOROE WALTER ROCHE, 

SECOND ASSISTANT EXCflXEER IN THE F. 8. NAVY. 



STEREOTYPED FOR THE AUTHORS, 

BY MEARS AND DUSENBERY, PHILADELPHIA. 

18 64. 



dC^ 



'•\^ 



Entered, according to Act of Congress, in the year 1864, by 
MARTIN ROCHE, M. D., 

AND 

GEORGE WALTER ROCHE, U. S. NAVY, 

in the Clerk's office of the District Court of the United States, in and for the Eastern 
District of Peansylvania. 



2. i> rsn 



WEMORY OF 



MAETIN KOCHE, 



PHILADELPHIA, 
FORMERLr 

Professor of Mathematics in the U, S. Navy, 

THK rOLLOWme pages, EXEMPUFYING his Km.ES FOR PRACTICAL NAVWATION, 

BEING THE SIMPLEST EVER DEVISED, ARE AFFECTIONATELY 
DEDICATED, BT HIS SONS, 



THE AUTHORS, 



(iii) 



PREFACE. 



Maiht persons wlio enter tlie maritime profession do not possess 
any knowledge of the ordinary rules of compntation ; otliers, again, 
begin tlieir career of following the sea, sufficiently prepared to 
perform any rule or method of calculation, but who from the nature 
of a seafaring life, do in the course of a few years, forget their early 
qualifications, and become altogether incompetent to solve even a 
question in Plane Sailing. 

The mariner, therefore, is in want of such practical methods as are 
most suitable to his calling, and most consistent with his limited 
acquirements. 

To supply the seaman's deficiency in this respect, and to facilitate 
the acquisition of knowledge so important as navigation, have been 
the Authors' aim and study for years, and it has been found that the 
best method by which the desired object can be accomplished, is to 
simplify the rules, so as to enable him without difficulty, to arrive at 
the greatest possible accuracy in the results. 

As the clearest and most convincing proof of the correctness and 
infallibility of these rules, they have worked out in full all of the 
examples in Bowditch's ISTavigator, and added Cases and Examples 
which that author has omitted to give in his "Epitome." 



(V) 



SIGNS USED IN THIS WORK. 



X 

{)■' 
V 



is the sign of Addition. 

*♦ " *' " Subtraction 

" " *' *' Multiplication, 

" " " " Division. 

" " " " the square. 

" " " the square root. 

" " " " equality. 

" " '* " a decimal, 

** *' *' " is to, in proportion. 

" *' *' " so is, in proportion, 

" " " " to, in proportion. 

" " " " a degree. 

" " *' " a minute of a degree. 

" *' '* " a second of a degree. 



EXPLANATION. 

The reason why the answers in the foot-notes to Examples of Cases IV. and V. of 
Plane Sailing differ from those worked out by the Table of Miles is, that the table is 
extended to only five places of decimals, and the fifth is calculated to the nearest 
figure. This applies to the answers in the foot-notes of Cases IV., V., VL, and VII. 
of Middle Latitude Sailing, and all the Cases of Mercator's Sailing. 



(vii) 



ARITHMETIC 



DECIMAL FRACTIONS. 

When any quantity is considered as constituting a whole, it is called an integer, 
and when an integer is supposed to be divided into a certain number of equal parts, 
any number of these parts, considered in their relation to the whole, is^ called a 
fraction, which is expressed by two numbers placed one above another, with a line 
between them in the form of a common or vulgar fraction ; the lower of these, called 
the denominator, denotes the number of parts into which the integer or whole one 
is divided, and the upper, called the numerator, expresses the number of these parts 
which is contained in the fraction : for instance, suppose a mile divided into 6 equal 
parts, and 2 of these parts were to be considered as a fraction of the whole ; it 
would be written in numbers thus, |, where the figure under the line shows that the 
mile is divided into 6 equal parts, and the 2 above the line denotes the number ol" 
those parts contained in the fraction. 

Fractions are of two kinds. Vulgar and Decimal. Decimal fractions vary in the 
game proportion, and are managed by the same methods of operation, as whole num- 
bers. For this purpose every proper fraction is reducible to another whose denomi- 
nator shall be 10, 100, 1000, &c. : viz. unity with some number of ciphers annexed ; 
and fractions with such denominators are called decimal fractions, such as /j, yVs^ 

As the denominator of a decimal fraction is always 10, or 100, or 1000, &c., the 
said denominators need not be written, for the numerators may be made to express 
the true value of a decimal. For this purpose it is only required to write the nume- 
rator with a point, called the separatrix, before it, to distinguish it from a whole 
number, when it consists of as many figures as the denominator has ciphers annexed 
to unity ; so y^^ may be written .5 ; yV^ may be written .75 ; and j%%^j^ may be writ- 
ten .625. But if the numerator has not so many places as the denominator has 
ciphers, put as many ciphers before it, viz. to the left hand, as will supply the defi- 
ciency ; so write j(yf .05 ; y^^jf .005. And thus do these fractions receive the form 
of whole numbers. 

Therefore we may consider the separatrix as the fixed point whence whole num- 
bers proceed to the left, infinitely increasing, and decimals to the right, infinitely 
decrea-?ing towards 0, as the following table shows : — 









o 'S § 'S • «• '^ 'S § ^ -« .2 

7654321 . 234567 

In decimals, as well as in whole numbers, each figure takes its value by its dis- 
tance from the separatrix ; if it be in the first place after the separatrix, it signifies 



2 ARITHMETIC. 

tenths ; if in the second, hundredths, &e., decreasing in each place to the right in a 
tenfold proportion. Consequently every single figure expressing a decimal has for 
its denominator 1, with as many ciphers as its place is distant from the separatrix. 
Thus 2 in the above table is /g, 3 is jgjj, and 4 is yjy'rtjj, &c. And if a decimal be 
expressed by several figures, the denominator is 1, with as many ciphers as the last 
figure is distant from the separatrix. So .234 signifies and must be read -^^^q. 

A cipher (or ciphers) placed at the right hand of a decimal fraction alters not its 
value, since every significant figure continues to occupy the same place ; so .5, .50, 
.oOO, &c., are all of the same value. 

But a cipher (or ciphers) placed to the left hand of a decimal does alter its value, 
every cipher decreasing it to j'jj of the value it had before, by removing every signi- 
ficant figure one place farther from the separatrix ; so .5, .05, .005 all express dif- 
ferent decimals, viz. .5, /^ ; .05, j§j ; .005, tuVs- 

"We may likewise observe the contrary efiect of ciphers being annexed to whole 
numbers and decimals ; every cipher to the right hand of a whole number increases 
its value ten times, but ciphers to the right hand of a decimal do not alter its value. 
Again, ciphers put to the left hand of a whole number do not alter its value, but 
every cipher put to the left hand of a decimal decreases its value to the j'j of what 
it would be without them, thus : — 



5. 


.5 


5. 


.5 


50. 


.50 


05. 


.05 


500. 


.500 


005. 


.005 


5000. 


.5000 


0005. 


.0005 



It is also manifest from the above table that the places of decimals decrease in a 
tenfold proportion from the separatrix towards the right hand, consequently they 
increase in a tenfold proportion towards the left hand, as the places of whole num- 
bers do ; for ten-hundredths make one-tenth, and ten-tenths make a unit ; ten units 
make ten, ten tens make one hundred, &c. ; viz. j^q = j'^, |g = 1 ; 1 X 10=^ 10, 
10 X 10 =100, &c., which proves that decimals are subject to the same law of nota- 
tion, and consequently of operation, as whole numbers are. 

Decimal fractions of unequal denominators are reduced to one common denomina- 
tor, when there are annexed to the right hand of those which have fewer places as 
many ciphers as make them equal in places with that which has most; so these 
decimals .5, .04, .125 may be reduced to the decimals .500, .040, and .125, which 
have all 1000 for their denominator. 

Of decimals that is the greater whose highest figure is greater ; whether they 
consist of an equal or unequal number of places : thus, .575 is greater than .395, 
and .5 greater than .395, for if it be reduced to the same denominator with .395 it 
will be .500, which is manifestly greater. 

A mixed number, viz. a whole number with a decimal annexed, is equal to an 
improper fraction, whose numerator is all the figures of the mixed number, taken 
as one whole number, and the denominator that of the decimal part ; so 32.405 is 
equal to YoVff » as is manifest from the method, given in all arithmetics, of reducing 
a mixed number to an improper fraction ; for 32, the integral part, being multiplied 
by 1000, the denominator of the fractional part, produces 32000, to which adding 
405, the numerator of the fractional part, the sum 32405 is the numerator to 1000 
for an improper fraction equal to the given mixed number. 

A pure whole number is one which has no decimal annexed to it. 
A pure decimal is one which has no whole number prefixed to it. 
A mixed decimal is one which has a whole number prefixed to it. 

ADDITION OF DECIMALS 

Is performed by writing down the given numbers to be added together so that the 
decimal points will be directly under each other (and if joined together would form 
a perpendicular line), and the figures will then be ranged in the proper order 
of their local values, viz. tenths under tenths, hundredths under hundredths, 
thousandths under thousandths, &c., and units under units, tens under tens, 
hundreds under hundreds, thousands under thousands, &c. ; then adding them up 
as if they were pure whole numbers, and in their sum placing the decimal point 
directly under the other decimal points in the given numbers. 







ARITHMETIC. 


d 


EXAMPLE I. 








EXAMPLE II. 




Add together 463., .422, 
and .914 

463. 
.422 
8.51 
^.43 

'.914 


8.51, 


.43, 


7., 


Add together .000342, 267., 
.45, and 402. 

.000342 
267. 
.826 
3.5 
.45 
402. 


.826, 3.5, 


Answer 480.276 


Answer 673.776342 





SUBTRACTION OF DECIMALS 



Is performed by writing down the given numbers to be subtracted, the less from the 
greater, so that the decimal points will be directly under each other (and if joined 
together would form a perpendicular line), and the figures will then be ranged 
in the proper order of their local values ; then subtracting them as if they were 
pure whole numbers, and in their difference placing the decimal point directly 
under the other decimal points in the given numbers. 



EXAMPLE L 
From 716048. subtract .0000625 
716048. 

.0000625 

Answer 716047.9999375 



EXAMPLE XL 
From 456321.23456789 subtract 42.3 
456321.23456789 
42.3 



Answer 456278.934561 



MULTIPLICATION OF DECIMALS 

Is performed by writing down the given numbers to be multiplied, as if they were 
pure whole numbers, and in their product placing the decimal point as many places 
from the right hand towards the left as there are decimal places in both the multi- 
plicand and multiplier together. 

N. B. — It sometimes happens that, in the multiplication of decimals, there will not 
be in the product as many figures as the rule requires decimal places, and in such 
cases the deficiency must be supplied by prefixing as many ciphers as there are 
places wanting. 



EXAMPLE L 
Multiply 34.26 by .542 
34.26 
.542 



6852 
13704 
17130 



18.56892 Answer. 



EXAMPLE II. 

Multiply .03526 by .00013 
.03526 
.00013 



10578 
3526 



.0000045838 Answer. 



Note. — To multiply by 10, 100, 1000, &c., remove the decimal point in the multi- 
plicand to the right as many places as there are ciphers in the multiplier. 

EXAMPLE. 

.426 X 10 = 4.260 ) 
.426X100 = 42.600 I Answer. 
.426X1000 = 426.000] 



ARITHMETIC. 



DIVISION OF DECIMALS 



Is performed by writing down the given numbers to be divided, as if they were pure 
whole numbers, and in the quotient placing the decimal point as many places from 
the right hand towards the left as there are decimal places in the dividend exceed- 
ing those in the divisor. This is the general rule ; there are, however, several par- 
ticular cases, which we shall treat of separately in order. 



EXAMPLE I. 
Divide 5.73 into 258.0219 

5.73)258.02' 19(45.03 Answer. 
2292 



2882 
2865 



1719 
1719 



EXAMPLE n. 
Divide 270.2 into 1275.6142 

270.2)1275.6'142(4.721 Answer. 
10808 



19481 
18914 



5674 
5404 

2702 
2702 



CASE I. 

When the number of places of decimals in the divisor and dividend is equal, the 
quotient will be a pure whole number, provided the division terminates evenly ; but 
if the division does not terminate evenly, and it is deemed necessary to annex 
ciphers to the dividend, then the quotient will be either a mixed or pure decimal 
having as many places as ciphers were annexed. 



EXAMPLE I. 

Divide 52.475 into 18208.825 

52.475)18208.825'(347. Answer. 
157425 



246632 
209900 

367325 
367325 



EXAMPLE IL 

Divide 23.8457 into 5365.3303 
23.8457)5365.3303 '(225.002004-f Ans. 
476914 



596190 
476914 

1192763 
1192285 



478000 
476914 



1086000 
953828 

132172 

N. B. — Five places of decimals will be 
suflficient for all calculations in practical 
navigation. 



Divide 53.497 into 42.968 



EXAMPLE nL 
Answer .803184- 



CASE II. 

When the number of places of decimals in the dividend exceeds that in the divi- 
sor, the quotient will be either a mixed or pure decimal, the^ number of places in 
which will be equal to the excess, provided the division terminates evenly ; but if 
the division does not terminate evenly, and it is deemed necessary to annex ciphers 
to the dividend, then the quotient will have as many places more as ciphers were 
annexed. 



ARITHMETIC. 



EXAMPLE I. 


Divide 314.52 into 1339.8552 


S14.52)1339.85'52(4.26 Answer 
125808 


81775 
62904 


188712 
188712 



EXAMPLE XL 
Divide .7402 into 19.73451 
.7462) 19.7345 ' 1 (26.44667+ Answer. 
14924 



48105 
44772 

33331 

29848 



Divide 8.2467 into 3.521698 



EXAMPLE III. 
Answer .42704-j- 



34830 

29848 

~49820 
44772 

50480 
44772 

57080 
52234 

4846 



CASE III. 

When there are not as many places of decimals in the dividend as there are in 
the divisor, ciphers must be annexed first to make them equal, and the quotient will 
be a pure whole number, provided the division terminates evenly ; but if the divi- 
sion does not terminate evenly, and it is deemed necessary to annex more ciphers to 
tlie dividend, then the quotient will be either a mixed or pure decimal having as 
many places as more ciphers were annexed. 

EXAMPLE L 

Divide 26.82345 into 9119.973 

26.82345)9119.973001(340. Answer. 
8047035 



10729380 
10729380 



EXAMPLE IL 
Divide 3456.782105 into 1451848.486 

3456.782105)1451848.486000' (420.00000054+ Answer. 

13827128420 



6913564400 
6913564210 



Divide 584.29685 into 26.4953 



19000000000 
17283910525 

17160894750 
13827128420 

3333766330 



EXAMPLE III. 
Answer .04534+ 



CASE lY. 

When the divisor is a pure whole number, and the dividend a mixed or pure deci- 
mal, the quotient will be either a mixed or pure decimal having as many places as 



6 ARITHMETIC. 

the dividend, provided the division terminates evenly ; but if the division does not 
terminate evenly, and it is deemed necessary to annex ciphers Uj the dividend, then 
the quotient will have as many places more as ciphers were annexed. 



EXAMPLE I. 
Divide 962. into 3354.625 

962.)3354.625(3.48713+ Answer. 

2886 



4686 
3848 

~8382 
7696 



6865 
6734 

1310 

962 



3480 

2886 

~594 



EXAMPLE XL 

Divide 5263. into 17.4269 

5263.)17.4269(.00331+ Answer. 
15789 

"16^579 

15789 

5900 
5263 

637 



EXAMPLE TIL 
Divide 74693. into .76829 Answer .00001028-|- 

Note. — To divide by 10, 100, 1000, &c., remove the decimal point to the left as 
many places as there are ciphers in the divisor. 

EXAMPLE. 

Divide 10 into 4.26 Answer .426 

" 100 " 42.6 " .426 

" 1000 " 426 " .426 

To divide by 60, remove the decimal point to the left one place, because there is 
one cipher, and proceed as in short division. 

EXAMPLE. 
Divide 60 into 8133.53992 
6,0)813,3.53992 

135.558998+ = 135.559 Ajiswer. 

This refers to Example of Case I. Plane Sailing, 



REDUCTION OF DECIMALS. 

The reduction of a vulgar fraction to a decimal is performed by annexing to the 
numerator as many ciphers as may be deemed necessary, and dividing the denomi- 
nator ; the quotient will be the decimal required, which must always consist of the 
same number of places as there were ciphers annexed. 

EXAMPLE II. 
To change .50041 of a degree to min- 



EXAMPLE L 
Reduce 30^ of a degree to a decimal. 

30° 
6,0)3,0. 

.5 Answer. 



30^ = 



This refers to Example of Case I. Par- 
allel Sailing, under "Remark." 



utes. 



Rule.— Multiply by 60, and point off 
as many places in the product as in the 
given decimal. 

.50041 
60 



30.02460 Answer. 



This refers to Example of Case II. Par- 
allel Sailing, under " Remark." 



ARITHMETIC. 



INVOLUTION 



Is the finding of the po-svers of numbers. The first power is the root ; the second 
power or square, is the product of the first power or root, multiplied l)y itself; the 
third power or cube, is the product of the square multiplied by its root, &c. 



EXAMPLE I. 

What is the square of 244? 

244 X 244 = 59536. Answer. 

This refers to Example of Case IV 

Plane Sailing. 



EXAMPLE II. 
What is the square of 136 ? 
136 X 136 = 18496. Answer. 
This refers to Example of Case IV. 
Plane Sailing. 



EXAMPLE in. 

What is the square of 203 ? 
203 X 203 = 41209. Answer. 
This refers to Example of Case V. Plane Sailing. 

EVOLUTION 

Is the finding of the roots of numbers, and of course, is the reverse of Involution. 
The root of a number is a factor, which being multiplied by itself a certain number 
of times, will produce the given number. Thus, if the given number is a square, 
the root is called the square root, if a cube, the root is called the cube root, &c. 

TO EXTRACT THE SQUARE ROOT. 

1st. Separate the given number into periods of two places, beginning at the sepa- 
ratrix, going to the left in whole numbers, but to the right in decimals. If there 
should be an odd figure left in whole numbers, it must be used as a full period ; but 
if there should be an odd figure left in decimals, it must not be used until made a 
full period by the addition of a cipher. 

2d. Find the greatest square in the left hand period or figure, and write its root 
down in the quotient, as in division ; square this root, and put it under the left- 
hand period or figure, and to the right of the remainder bring down the next period 
in order, for a dividend. 

3d. Double the figure of the root already found, for a trial divisor, and find how 
often it is contained in the dividend exclusive of the right hand figure, and write 
the number in the quotient as the second figure of the root ; annex it also to the 
trial divisor, which makes the true divisor. Multiply the true divisor by the second 
figure of the root, and subtract the product from the dividend. To the remainder 
bring down the next period, for a new dividend. 

4th. Double the figures of the root already found, for a trial divisor, and proceed 
as before until ail the periods have been brought down. 

N. B. — Every period of the given number must give a figure in the root. 

If the given number be a pure whole number, and the extraction of the root ter- 
minates evenly, the root will contain as many places of whole numbers as there were 
periods in the*^ given number; but if the extraction does not terminate evenly, and 
it be deemed necessary to annex periods of decimals, then the root will contain as 
many places of decimals as there were periods annexed. 

If the given number be a mixed decimal, and the extraction of the root terminates 
evenly, the root will contain as many places of whole numbers as there were periods 
in the given whole number, and as many places of decimals as there were periods 
in the given decimal ; but if the extraction does not terminate evenly, and it be 
deemed necessary to annex more periods of decimals, then the root will contain as 
many more places of decimals as there were periods annexed. 

If the given number be a pure decimal, and the extraction terminates evenly, the 
root will contain as many places as there were periods in the given decimal ; but if 
the extraction does not terminate evenly, and it be deemed necessary to annex more 
periods of decimals, then the root will contain as many more places of decimals as 
there were periods annexed. 



ARITHMETIC. 

EXAMPLE I. 

Extract the square root of 41040. 

4|l0|40.(202.583+ Answer. 
4 



402)1040 
804 



4045)23600 
20225 



40508)337500 
324064 



405163)1343600 
1215489 



128111 



This refers to Example of Case IV. Plane Sailing. 

EXAMPLE XL 

Extract the square root of 18327 
l|83|27.(135.377+ Answer. 

23 ]83 
69 



265)1427 
1325 



2703)10200 
8109 



27067)209100 
189469 



270747)1963100 

1895229 



67871 



This refers to Example of Case V. Plane Sailing. 



EXAMPLE IIL 

Extract the square root of .17977 

.17'97l70(.42399+ Answer. 
16 



82)197 
164 



843)3370 
2529 

8469)84100 
76221 



84789)787900 
763101 

"^4799 



1 


o .c rs 3-. 

— r^ OS M s i~ '3 

"^ ffl «5 O us O .O 


pi .~r '"- i"- ■;: - 


s H^ r ■=■ ?: '.: 

^ '^ CC M CO M 


•c .- o, a: oc =5 '- t * ^' -* -5 
?i c4 01 ^^■ oi oi m rn — ■ i-. rn ^ 


-t « ^ 35 » CO 

I - i is CO Ji 

r-; =5 C3 d d C3 


2 


O r-c ^( r3 TT O 


oooo = o 
o ri :^ .■'5 Tt io 

s 


§2S^^5 


82?, .^^g 


g2§s^5 


35 


a 


■* ^^ T-i ■>! — : — — ^ 


mm 


iiiill 

o o o 3; 35 rs 


g i5 ::: 2 g ?? 
So.3S8S 

35 d 3: 00 3C od 




:;• ^ :^ 2 ;'' !(S 

I- i- d d d X) 


3 




S2^^^5 

33 




g2^^^5 


825^55 


g2Sg2g 

CO 1 


c 

I"" 




1, I, ,,: i~ -si d 


S =^. ^1 = =<> =^ 

;0 ;i -.S •.-- 'C. o 




S g i £ 1 1=2 
^ ^ ^ p ^ ^. 

-f -f -f •* coco 


gg2Sd^ 

ri. c; r-l 35 iO 

CO 2 2 :i^' 2 2 




s~sr^5s 


g2§g^g 


82§§5g 
•-0 


g2i^?g 


bi; 

1 


liiiiii 


S^ ^ ?5 ?) fl ?i 


R2^f KH 


05 o5 oj o3 oi ?i 


S 55 s ?, 2 2 








§2Sg^g 


S2§g^3 


825^^5 
s 


8 ;::; oi c^o ? 5 
o 


g2Sg5g 


ib 


liiiiii 

?? ?i ?i ^ ?i ?5 




iiiill 


iiiiii 


O -r j£ CO j-^ ^ 

?, fj ol 05 o1 ^1 


^^^;^k^. 


"S 




52§g^S 


§2g^^S 82^^55 


82^^i5 


8238^8 




^ ^ ;? ;? s M 
°3 


>ii « 5 -* S ;:3 
O ^ C-J CO '3' o 


i? So ?o ?i fc ?o 

O rt -M CO -C O 


O .-^ 0^ CO -T O 


S 2 i i g ^ 

4il-fl-i- 

O i- O, CO -f o 


8 S 2 5 8 S 
8 8 c^ 8 5 8 


5 ;^ s CO ?■ 


ii 

3 


liiii 




CC » 30 OC OC l^ 

CO CO ?0 CO CO CO 


?:?^Sg2K 

t~ O ^ CO 05 O 


iiiiii 


Iiiiii 

■4 is i2 15; ^' 1? 


^ 




S2?,g5S 


52^M^g 


g2§§^S 


g2^i^§ 

05 


820-58^8 

CO 
•0 




ii! 




3 § S A ?i S 

235;£s:^ 

CO CO 05 oi oi o4 


S S5 i? :^ S 8 

Iiiill 

■^' -+ -^ ^ ^ ^! 


q o ^ S ^ q 


mm 

iiiiii 


3 


° CM 


§ 2 s -^ 5 s 

CO 

-1" 


§2S^^g 


82^^^S 




8288^8 


bi 


. 5 aJ § :? s '^ 

liiiiii 

" 3C OC 2 =«' » ^ 


-2S^^S 


iiii 




S o ^ ^: o s 

■i o ^? -^ ^ o 
03 03 t- n o CO 
o .«■ O O O uO 


2^.2§SS2 

uO -l« -t -* 
-f -f -f ^T)-^ 


3 




§2Sg^§ 


S2§g^g 


g2S^^S 


oooooo 

Or-105 CO-* O 
O 


OOOOOO ' 
r-c 05 rC .0 


ti 

c 


iifiiii 


o 2 2 ^ S "^ 
rf CO oq ^_ q S 


S85ggg=: 


CO 05 — O OCC 




l-H 35 OC I-- 


o o ..-: .o lO .« 
o ~, 


■17 O »0 O >0 lO 
Or^ <N CO -* o 


§2g^iS 

oi 


82Sgi§ 


oooooo 
Oi-i !N contio 


©2558^8 


1 


liiiiii 

r 00 1~ » o u- -* 

iC O O O lO kO 


Siiili 

vO o o o .o o 


ligiii 

lisi§i 


-+ CO co_ 05 ^_ q 

CO co' CO' CO' CO' CO 
lO O O O uO o 


iiiiii 

O o i^ ^ o o 


(M" 05 05" 05' 05 0<i 

iC vo UO 


1 


~ §2§i§S 




O — OJ CO-* o 


oooooo 

O r-l <M CO Tt iO 


82g,g^g 

25 


82§8^g 


1 


ipiiiis 

S 5 5 .1 o 5 


o o -* >o o o 


§ § S S o o 




iiiiii 


iSiiii 


3 




i2§S^S 

C5 




g2§^5S 


g2Sg^g 


oooooo 

r- C^ CO ^ 
05 


i r 

►3 


S -o ^- S O uO S 




CC CO CC OC OC CO 

.o o UO O iC o 


q q =q co i, t^ 


t^ M ■* O, .-0 O 

q q o ir ^ t, 
i^ t-l i^ i~^ t-^ t~ 

UO .O O uo uO UO 


2SS§sS 

CC uO CO t- 

5 5 5S5S 


i 




S2SS5S 

CO 


g2Sg5S 


g2S^5S 


82Sg§S 
2 


g2S8§g 






O uO O -f -t -I* 


o o -i< oc bi CO 

2 ^ ;-2 ? 2 5 

-* CO CO CO CO o? 

§ s S g o S 


o5 ?5 ^ J-; ;2 iii 


^ § ?^ ^1 ;j s 

§?5^§gg 

q q q q q q 

35 35' 35 3C' 00' 00 




s 




§2§g^S 


g2§^§§ 

00 


82gg^g 

05 


g2§g^g 



52^852 


1 ^ 




iiiiii 


iiisii 




iiiiii 

iSiSig 


gggggs 




°o 


§2§g^g 




gsss^s 

to 


82§§§S 


g2§8^S . 



(9) 



o 









I 






?3i 



o 
9 

ll 
!^ 

1 


S'! ?^' ?i 


www 


aaaa 
aaaa 

aaa 


b i? it is 
y.yyy. 

aaaa 


a 


1 

u 
r 

o 


i 

J= a 

•3 
g 




^•^'^'^ 

J = = ^ 

r.Kss 
s =^ =4 =^' 

CfcOJ CO 


a a a a 
^■aaa 

a'a::^ 

'/. X X 


> > > >^ 
aixx X 

aaaa 
* M oi X 


o 

V2 


o 

2 
K 

5 

3 
W 


Miles in 1° of 
Longitude in 
Lutitude equal 
to the Coniple. 
uient of the 
Cour.so. 


d :n 'ooc 


i^ o ^ ?5 


iiii 


2gS^ 
^ '- ==. -^ 


■^; 


Iiii 

Ills 


ill 




30 u-JM O 






i 


^ 1 

11 1 

^! 1 


1? A 


oci^r,';L* 


t-S'i'i* 


■^'szrj; 


o?"^"^* 


-* 




iiii 

q o. i^ cc 


F^ S a ^ 


ui O O >0 


22gg 


?3 

-* 


Iflll- 


i '1 


nH— let- 1» 

o O <N O OC 




g2sf 






° 1 

pi 


o 

1 

3 

li 
II 

■-•3 
o 

i 
1 


o = s o 


^^'--'^T 


c<) 5rt^'?r 


cc S:*?:'^ 


^ 


aiaioo 


b >1 >^ >5 

crJxx 


-^. X X X 
^- a; X X 

fe. '3 g ^ 

^ si a a 

^^aa-a 

aixx 

iXXX 


M . . . 

^^■^^ 
X X X X 

aiaixai 

-aaa 

X 




1 

li 

= 1 

3rS 
1 

5 

3 


1 

3 

ft 

^ -J 
1 

i 


52ZZ 


b >-. >-. >■• 

- a a a 

Jsaa 

^2Z 


= c c c 

a s a s 

Zaas 


^■^^^ 

zyyy. 
yy.yy. 

J22Z 


z 

o 

a 

z 


1 
1 

1 

Ed 

1 



,.2 



X 

C W 



c -s 

X ^ 

J; O 

6 J 



^ H 









CIO) 



^U'S 



^04r 




The Mariner's Compass is an artificial representation of the horizon of every 
pla<;e. It consists of a circular card divided into 32 equal parts, by lines drawn 
from the centre towards the circumference, called rhumb-lines, the extremities of 
which are called points ; the intervals are subdivided into half and quarter points. 
The points of the compass are counted from the meridian or the North and South 
points towards the East and West points of the horizon, and are named as follows: — 



Points. 


North-East Quarter. 


North-West Quarter. 


South-East Quarter. 


South-West Quarter. 





North 


North 


South 


South 


1 


N. by E. 


N. by W. 


S. by E. 

S.S.E. 


S. by W. 


2 


N.N.E. 


N.N.W. 


s.s.w. 


3 


N.E. by N. 


N.W. by N. 


S.E. by S. 


S.W. by S 


4 


N.E. 


N.W. 


S.E. 


s.w. 


5 


N.E. by E. 
E.N.E. 


N.W. by W. 


' S.E. by E. 


S.W. by W. 


6 


W.N.W. 


E.S.E. 


w.s.w. 


7 


E. by N. 


W. by N. 


E. by S. 


W. by S. 


8 


East 


West 


East 


West 

,1 



(11) 



12 MARINER'S COMPASS. 

The four principal points are called Cardinal Points, two of which, opposite each 
other, are called North and South points ; that which is on the right hand when 
you look towards the North is termed the East, and that which is on the left hand, 
or opposite, the West point. The names of the other points are compounded of these, 
according to their situation or position, but instead of the words the initials only are 
used. The North, South, East, and West are usually marked vrith a single letter, 
thus : N. S. E. and W. The North point is frequently marked with the Fleur de Lis. 
The North and South are markefl because they are on the meridian. The 
4 points are expressed by two letters. The 2 and 6 points are expressed by three 
letters. The remaining 16 points are marked with the word bi/ in their names. The 
3 and 7 points are always named and read backwards from the 4 and 8 points, but 
counted forward from the meridian, as all the others are. 

Under the card, along the North and South line, a small bar of steel is fixed, 
called the Needle, which being magnetized, or touched with a loadstone, acquires 
the peculiar property of pointing north and south, and consequently by the card, 
determines the direction of the other points of the horizon. The needle having a 
small socket in the centre, is supported, together with the card, on the point of a fine 
steel pin, on which it freely turns, and by the above-mentioned property, its points 
keep always in the same direction (the compass is here supposed to be true). These 
are confined in a circular brass box with a glass cover, the box being hung in brass 
hoops or gimbals, in order to counteract the motion of the ship. The whole of these 
are placed in a square wooden box with a moveable lid, which serves to support the 
gimbals and secure the compass from accident in removals. 

The compass is used to point out the direction that a ship sails at sea. For this 
purpose, it is so placed in the ship that the middle section of the wooden box, par- 
allel to its sides, may be parallel to the middle section of the ship along its keel. 
When it is thus fixed, that point of the card which coincides with a perpendicu- 
lar line marked on the inside of the circular box, and termed by seamen the 
'^Lubber's Faint/' will show the direction of the ship's head. 

To " box the compass,'' or, in other words, to name the various points of it accu- 
rately, is the first, the easiest and the simplest lesson in navigation. Many persons 
imagine this to be so puzzling, that they give it up at once and for ever, but the 
whole task should be accomplished in about half an hour. So begin with mastering 
one quarter of the compass, and the rest is easily accomplished, merely by a change 
of name, for the same arrangement applies to the other three quarters. We will 
now give two very simple rules by which you may learn the whole " mystery.'' 

KuLE I. — When the letters indicating two points are united, the point meant is 
halfway between the two; thus, N.E. is halfway between North and East; N.N.E. 
is half way between North and N.E. ; E.N.E. is half way between East and N.E. 

KuLE II. — When the letters are joined by the word by or the letter b, the point 
meant is the one which comes next after the first going towards the second ; thus, 
N. by E. is next to North going East ; N.E. by N. is next to N.E. going North ; 
N.E. by E. is next to N.E. going East; and E. by N. is next to East going North. 



ON THE TABLE OF MILES IN A DEGEEE OF 
LONGITUDE. 



CASE I. 

GiTEN the latitude to find the corresponding number of miles in a degree of lon- 
gitude. 

RULE. 

Find, in the Table of Miles, the next less and the next greater than the given 
latitude, and take out the corresponding miles in a degree of longitude. One-tenth 
of the difference between these will be the difference in miles for one minute in the 
latitude. Multiply this difference for one minute by the number of minutes the 
given latitude exceeds the next less, and subtract the product from the number of 
miles in a degree of longitude corresponding to the latitude next less than the given 
one, and you will have the number of miles in a degree of longitude in the given 
latitude. 



EXAMPLE L 

Required the number of miles in a de- 
gree of longitude in latitude 33° 52^ N. 
orS. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

. . 33^50^ = 49.83966 



Next less 
" greater 



34 00 =49.74218 





10 

. 33° 50^ = 

-f 02 = 

. 33 52 = 


). 09748 






.009748 
2 

^19496 




Next less . . 
Given latitude . 


= 49.83966 
= — .01950 

= 49.82016 


Next less . 
Given latitude 



This refers to Example of Case IV. 
Plane Sailing. 



EXAMPLE IL 

Required the number of miles in a de- 
gree of longitude in latitude 56° 18^ N. 
or S. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

Next less . . . 56° 10^ = 33.40677 
, 56 20 =33.26160 



greater 



10 ). 14517 

.014517 
8 

.116136 

56° 10^ = 33^0'677~ 
4-08 = — .11614 

56 18 = 33.29063 



This refers to Example of Case V. 
Plane Sailing. 

(13) 



14 



ON THE TABLE OF MILES IN A DEGREE OF LONGITUDE. 



EXAMPLE IIL 

Required the number of miles in a de- 
gree of longitude iu latitude 28° iV N. 
or S. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

Next less . . . 28° 10^ = 52.89465 
" greater . . 28 20 =52.81207 



Next less 



Given latitude 



10 



28° 10^ 
+ 01 



). 08258 

.00^^258 

52.89465 
— .00826 



28 11 



52.88639 



This refers to Question IV. for exercise 
in Plane Sailing. 

EXAMPLE IV. 

Required the number of miles in a de- 
gree of longitude in latitude 25° 04^ N. 
orS. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

. . 25° 00^ = 54.37850 



Next less 
" greater 



25 10 =54.30438 



10 ).07412 

.007412 
4 


Next less 
'* greater 


.029648 

25° 00^ = 54.37850 
+ 04 = — .02965 


Next less . 


25 04 = 54.34885 


Given latitude 



Next less 



Given latitude 



This refers to Question V. for exercise 
in Plane Sailing. 

EXAMPLE V. 

Required the number of miles in a de- 
gree of longitude in latitude 53° 36-' N. 
or S. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 



Next less . . 
" greater . 


. 53° 30^ 
. 53 40 


= 35.68934 
= 35.54886 


Next less 
" greater 




10 


). 14048 

.014048 







.084288 




Next less . . 


. 53° 30^ 
+ 06 


= 35.68934 
= — .08429 


Next less . 


Given latitude 


. 53 36 


= 35.60505 


Given latitude 



This refers to Question II. for exercise 
in Parallel Sailing. 



EXAMPLE VL 

Required the numljer of miles in a de- 
gree of longitude in latitude 32° 17^ N. 
or S. 

By the Table of Miles in a degree of 
hjngitude, &c., on page 9, 



Next less 
" greater 



32° 10^ = 50.79023 
32 20 =50.69698 



Next less 



Given latitude 



10 



32° 10^ 

+ 07 



). 09325 

.009325 
7 

.065275 

o0T9'023" 
— .06528 



32 1^ 



50.72495 



This refers to Question III. for exercise 
in Parallel Sailing. 

EXAMPLE VIL 

Required the number of miles in a de- 
gree of longitude in latitude 39° 3F N. 

or S. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

39° 30^ = 46.29745 
39 40 =46.18628 



10 



). 11117 

xTiTin 



39° 30^ = 46.29745 
+ 01 = — .01112 



59 31 



46.28633 



This refers to Example of Case I. Mid- 
dle Latitude Sailing. 

EXAMPLE VIIL 

Required the number of miles in a de- 
o-ree of longitude in latitude 48° 38'' N. 
or S. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

48° 30^ = 39.75725 
48 40 =39.62627 



10 


). 13098 




.013098 
8 




.104784 


48° 30^ = 

+ 08 = 


= 39.75725 
= -.10478 


48 38 = 


= 39.65247 



This refers to Example of Case II. Mid- 
dle Latitude Sailing. 



ON THE TABLE OF MILES IN A DEGREE OF LONGITUDE. 



15 



EXAMPLE IX. 

Required the number of miles in a de- 
gree of longitude iu latitude -iC 26^ N. 
or S. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 



Next less 
'* greater 



40° 20^ = 45.73747 
40 30 := 45.62442 



Next less 



Given latitude 



10 ). 11305 

.011305 
6 

^07830 

40° 20^ = 45.73747 
4- 06 = — .06783 



40 26 



45.66964 



This refers to Example of Case III. Mid- 
dle Latitude Sailing/. 

EXAMPLE X. 

Required the number of miles in a de- 
gree of longitude in latitude 48° oV N. 
or S. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

Next less . . . 48° 50^ = 39.49509 
" greater . . 49 00 =30.36355 



10 



). 13154 
.013154 



Next less 



Given latitude 



48° 50^ = 39.49509 
+ 01 = — .01315 



48 51 



39.48194 



This refers to Example of Case IV. Mid- 
dle Latitude Sailinc/. 

EXAMPLE XL 

Required the number of miles in a de- 
gree of longitude in latitude 53° 08^ N. 
orS. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

, 53° 00^ = 36.10975 
53 10 =35.96933 



Next less 
" greater 



10 



). 14042 



.014042 



.112336 



Next less 



Given latitude 



53° 00^= 36.10975 
+ 08 = — .11234 



53 08 



35.99741 



This refers to Example af Case Y. Mid- 
dle Latitude and Mercator's Sailing. 



EXAxMPLE XIL 
Required the number of miles in a de- 
gree of longitude in latitude 50° 43^ N. 
or o. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 



Next less 
" greater 



50° 40^ = 38.02982 
50 50 =37.89465 



Next less 



Given latitude 



10 



50° 40^ 

+ 03 



). 13517 

.013517 
3 

.040551 

38.0l982~ 
— .04055 



50 43 



.9^927 



This refers to Example of Case VI. Mid- 
dle Latitude Sailing. 

EXAxMPLE XIIL 

Required the number of miles in a de- 
gree of longitude in latitude 50° 58^ N. 
or S. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 



Next less 
" greater 



Next less 



Given latitude 



50° 50^ ==37.89465 
51 00 =37.75922 

10 ). 13543 

1)13543 
8 

.108344 

50° 50^ = 37y9"465~ 
-f 08 = — .10834 

50 58 = 37.78631 



This refers to Example of Case VII. 
Middle Latitude and Mercator^s Sailing. 

EXAMPLE XIV. 

Required the number of miles in a de- 
gree of longitude in latitude 48° 23^ N. 
or S. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

, 48° 20^ = 39.88771 
, 48 30 =39.75725 



Next less 
" greater 



Next less 



Given latitude 



10 



48° 20^ 
+ 03 



). 13046 

^13046 
3 

.039138 

39.88771 
— .03914 



48 23 = 39.84857 



This refers to Example of Case VII. 
Middle Latitude Sailing. 



16 



ON THE TABLE OF MILES IN A DEGREE OF LONGITUDE. 



EXAxMPLE XV. 

Eequired the number of miles in a de- 
gree of longitude in latitude 35° H'' N. 
or S. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 
Next less . . . 35° 10^ = 49.04876 

** greater . . 35 20 =48.94798 



10 



). 10078 



.010078 



.070546 



Next less 



35° 10^ = 49.04876 
+ 07 = — .07055 



35 r 



48.97821 



Given latitude . 

This refers to Question I. for exercise 
in Middle Latitude Sailing. 

EXAMPLE XVL 

Required the number of miles in a de- 
gree of longitude in latitude 34° 47^ N. 
orS. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 



This refers to Question II. for exercise 
in Middle Latitude Sailing. 

EXAMPLE XVIL 

Required the number of miles in a de- 
gree of longitude in latitude 79° 56^ N. 
or S. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 
Next less . . . 79° 50^ = 10.59073 

" greater . . 80 00 =10.41887 



10 



).17186 



.017186 
6 



.103116 



EXAMPLE XVIIL 



Required the number of miles in a de- 
gree of lon";itude in latitude 10° 04^ N. 
or b. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 



<ext less . . 
" greater . 


. 10° 00^ = 59.08838 
. 10 10 =59.05795 




10 ). 03043 




.003043 
4 



Next less 



Given latitude 



.012172 

10° 00^ = 59ir8'83F 
+ 04 = — .01217 



10 04 



59.07021 



This refers to Question III. for exercise 
in Middle Latitude Sailing. 

EXAMPLE XIX. 

Required the number of miles in a de- 
gree of longitude in latitude 37° 13^ N. 
or S. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 



Next less . . 
*' greater . 


. 34° 40^ 3 
. 34 50 : 


= 49.34852 
= 49.24898 


Next less . . 
" greater . 

Next less . . 
Given latitude 


. 37° 10^ = 47.81216 
. 37 20 =47.70725 




10 

. 34° 40^ = 

+ 07 = 


).09954 


10 ). 10491 




.009954 

7 


.010491 
3 




.069678 


.031473 


Next less . . 


= 49.34852 
= — .06968 


. 37° 10^ = 47.81216 
+ 03 = — .03147 


Given latitude 


. 34 47 = 


= 49.27884 


. 37 13 = 47.78069 



This refers to Question III. for exercise 
in Middle Latitude Sailing. 

EXAMPLE XX. 

Required the number of miles in a de- 
gree of longitude in latitude 83° 36^ N. 
or S. 

By the Table of Miles in a degree of 
longitude, &c., on page " 



Next less . , 
" greater 



83° 30^ = 6.79219 
83 40 =6.61875 



Next less 



Given latitude 



79° 50^ = 10.59073 
+ 06 = — .10312 



79 56 



10.48761 



This refers to Question III. for exercise 
in Middle Latitude Sailing. 



Next less 



Given latitude 



10 



83° 30^ = 
+ 06 = 



).17344 

.017344 
6 

.104064 

6.79219 
- .10406 



83 36 



6.68813 



This refers to Question IV. for exercise 
in Middle Latitude Sailing. 



ON THE TABLE OF MILES IN A DEGREE OF LONGITUDE. 



17 



EXAMPLE XXL 

Required the number of miles in a de- 
gree of longitude iu latitude 6° 24^ N. 
orS. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 



Next less . 
*' greater 



6° 20^ = 50.63370 
6 30 =: 59.61425 



10 



1.01945 



.001945 
4 



Next less 



Given latitude 



.007780 

6° 20^ = 59.63370 
+ 04 = — .00778 



6 24 



59.62592 



This refers to Question IV. for exercise 
in Middle Latitude Sailing. 

EXAMPLE XXIL 

Required the number of miles in a de- 
gree of longitude in latitude 44° 39^ N. 
orS. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 



Next less 
" greater 



44° 30^ = 42.79500 
44 40 =42.67248 



10 



). 12252 

.012252 
9 



Next less 



Given latitude 



.110268 

44° 30^ = 42 J950F 
-f 09 = — .11027 



44 39 



42.68473 



This refers to Question V. for exercise 
in Middle Latitude Sailing. 

EXAxMPLE XXIIL 

Required the number of miles in a de- 
gree of longitude in latitude 38° 25^ N. 
or S. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

Next less .... 38° 20^ = 47.06489 
" greater . . 38 30 =46.95641 



2)76 50 94.02130 



Given latitude 



38 



47.01065 



"When the minutes are halfway be- 
tween two of the numbers in the Table, 
the arithmetical mean of the two may be 
taken, which will give the number of 
miles required, the same as by the above 
rule. 

This refers to Question VI. for exercise 
in Middle Latitude Sailing. 
2 



EXAMPLE XXIV. 

Required the number of miles in a de- 
gree of longitude in latitude 47° 44^ N. 
orS. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 



Next less 
" greater 



Next less 



Given latitude 



47° 40^ = 40.40657 
47 50 =40.27731 

10 ). 12926 

.012926 
4 

.051704 

47° 40^ = 4oTo'657" 
+ 04 = — .05170 



47 44 



40.35481 



This refers to Question VII. for exercise 
in Middle Latitude Sailing. 

EXAMPLE XXV. 

Required the number of miles in a de- 
gree of longitude in latitude 53° 28^ N. 
or S. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 



Next less 
*' greater 



53° 20^ = 35.82950 
53 30 =35.68934 



Next less 



Given latitude 



10 



53° 20^ 
+ 08 



). 14016 

"ioiioTe 

8 

.112128 

35^2'950' 
— .11213 



53 28 = 35.71737 



This refers to Question VIII. for exer- 
cise in Middle Latitude Sailing. 

EXAMPLE XX YL 

Required the number of miles in a de- 
gree of longitude in latitude 50° 49^ N. 
or S. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 
Next less . . . 50° 40^ = 38.02982 

*' .greater . . 50 50 =37.89465 



Next less . 



Given latitude 



10 


).13517 




.013517 
9 


50° 40^ 

+ 09 


.121653 

= 38.02982 
= — .12165 


50 49 


= 37.90817 



This refers to Question IX. for exerci.se 
in Middle Latitude Sailing, and Question 
V. for exercise in Mercator's Sailing. 



18 



ON THE TABLE OF MILES IN A DEGREE OF LONGITUDE. 



EXAMPLE XXVn. 
Required the number of miles in a de- 
gree of longitude in latitude 46° 06^ N. 
or S. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 



Next less . . 
" greater . 


. 46° 00^ = 41.67942 
. 46 10 =41.55371 


Next less . . 
" greater . 

Next less . . 
Given latitude 


. 38° 50^ 
. 39 00 


= 46.73839 
= 46.62882 




10 ). 12571 

^12571 
6 

^7'5426 


10 

. 38° 50^ • 

+ 04 : 

. 38 54 


). 10957 

.010957 
4 




.043828 


Next less . . 


. 46° 00^= 41.67942 
+ 06 = — .07543 


= 46.73839 
= _ .04383 


Given latitude . 


. 46 06 = 41.60399 


= 46.69456 









This refers to Question IX. for exercise 
in Middle Latiiude Sailing. ' 

EXAMPLE XXVIIL 

Required the number of miles in a de- 
gree of longitude in latitude 38° 33^ N. 
or S. 

Qy the Table of Miles in a degree of 
longitude, &c., on page 9, 



Next less 
" greater 



38° 30^ 
38 40 



46.95641 
46.84753 



Next less 



Given latitude 



10 ).10888 

.010888 
3 

.032664 

38° 30^ = 46^64r 
+ 03 = — .03266 



38 33 



46.92375 



EXAMPLE XXIX. 

Required the number of miles in a de- 
gree of longitude in latitude 38° 54^ N. 
or S. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 



This refers to Example of Case II. Mer- 
caior's Sailing. 

EXAMPLE XXX. 

Required the number of miles in a de- 
gree of longitude in latitude 38° 55^ N. 
or S. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 



Next less 
" greater 



Next less 



Given latitude 



38° 50^ = 46.73839 
39 00 =46.62882 

10 ). 10957 

.010957 
5 

.054785 

38° 50^ = 46T3'839' 
+ 05 = — .05479 

38 55 = 46.68360 



It may be calculated as Example 
XXIII. 

This refers to Question I. for exercise 
in Mercator's Suiling. 



This refers to Example of Compound 
Courses at the end of Middle Latitude 
Sailing. 

EXAMPLE XXXL 

Required the number of miles in a degree of longitude in latitude 21° 52^ N.or S. 

By the Table of Miles in a degree of longitude, &g;, on page 9, 



Next less . . 
" greater . 


. 21° 50^ = 
. 22 00 ^ 


= 55.69615 
= 55.63103 




10 

. 21° 50^ 3: 

+ 02 = 


).06512 




.006512 

2 




.013024 


Next less . . 


= 55.69615 
= — .01302 


Given latitude 


. 21 52 = 


= 55.68313 



This refers to Example of Compound Courses at the end of Mercator^s Sailing, 



ON THE TABLE OF MILES IN A DEGREE OF LONGITUDE. 



19 



CASE II. 

Given the number of miles in a degree of longitude to find the corresponding 
latitude. 

RULE. 

Find, in the Table of Miles, the next less and the next greater than the given 
number of miles in a degree of longitude, and take out the corresponding latitudes, 
and take their difference. Also take the difference betvrcen the next less number 
of miles and the given one. Multiply this last difference by 10, and divide by the 
difference bet^veeu the next less and the next greater ; the quotient will be a num- 
ber of minutes to be subtracted from the latitude answering to the next less, and 
the renuiinder will be the latitude corresponding to the given number of miles in a 
desree of longitude. 



EXAMPLE L 

Required the latitude in which the 
number of miles in a degree of longitude 
is = 33.44262 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 
56° 00^ = 33.55163 Next greater 
56 10 =33.40677 " less 



10 



.14486 



Given No. 33.44262 
Next less, 33.40677 



.03585 
10 

).35850(2^ 
28972 



6878 



56° 10^ 
— 02 



33.40677 Next less 
+ .03585 



56 08 -^ 33.44262 Given number. 



This refers to Example of Case IV. 
Plane Sailing. 

EXAMPLE IL 

Required the latitude in which the 
number of miles in a degree of longitude 
is = 49.91803 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 
33° 40^ = 49.93655 Next greater 
33 50 =49.83966 " less 



10 



.09689 



Given No. 49.91803 
Next less, 49.83966 



.07837 
10 

).78370(8^ 
77512 



858 



83° 50^ = 49.83966 Next less 
— 08 = + .07837 



33 42 = 49.91803 Given number. 



This refers to Example of Case V. 
Plane Sailing. 



EXAMPLE IIL 
Required the latitude in which the 
number of miles in a degree of longitude 
is = 33.39527 

By the Table of Miles in a degree of 
longitude, &c,, on page 9, 

56° 10^ = 33.40677 Next greater 
56 20 =33.26160 " less 



10 



.14517 



Given No. 33.39527 
Next less, 33.26160 



.13367 
10 

)1.33670(9^ 
130653 



3017 



56° 20^= 33.26160 Next less 
— 09 = + .13367 



56 11 = 33.39527 Given number. 



This refers to Example of Case 
Plane Sailing. 



VI. 



EXAMPLE IV. 

Required the latitude in which the 
number of miles in a degree of longitude 
is =28.33333 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 
61° 40^ = 28.47601 Next greater 
61 50 =28.32227 " less 



10 



.15374 

Given No. 28.33333 
Next less, 28.32227 



.01106 
10 



.11060(F nearly. 
L5374 ^ 



61° 50^ 
— 01 



28.32227 Next less 
+ .01106 



61 49 



This refers to Question IV. for exercise 
in Plane Sailing, 



20 



ON THE TABLE OF MILES IN A DEGREE OF LONGITUDE. 



EXAMPLE V. 

Required the Latitude in which the 
number of miles in a degree of longitude 
is =25.42373 

Bj the Table of Miles in a degree of 
longitude, &c., on page 9, 

64° 50^ = 25.51515 Next greater 
65 00 =25.35708 " less 



10 = .15807 



Given No. 25.42373 
Next less, 25.35708 





.06665 
10 




).66650(4' 
63228 




3422 


65° 00^ 
— 04 


= 25.35708 Next less 
= + .06665 


64 56 


= 25.42373 Given number. 



This refers to Question V. for exercise 
in Plane Sailing, 



EXAMPLE VL 

Required the latitude in which the 
number of miles in a degree of longitude 
is = 52.13821 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

29° 30^ = 52.22133 Next greater 
29 40 =52.13518 " less 



10 = .08615 



Given No. 52.13821 
Next less, 52.13518 





.00303 
10 


29^ 40^ 
-00 


).03030 

= 52.13518 Next less 
= + .00303 


29 40 


= 52.13821 Given number. 



This refers to Question VI. for exercise 
in Plane Sailing. 



EXAMPLE VIL 

Required the latitude in which the 
number of miles in a degree of longitude 
18 = 58.62987 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

12° 10' = 58.65230 Next greater 
12 20 =58.61527 " fess 



10 = .03703 



Given No. 58.62987 
Next less, 58.61527 

.01460 
10 



(.14600(4'' nearly. 
14812 



12° 20'= 58.61527 Next less 
— 04 = + .01460 



12 16 = 58.62987 Given number. 



This refers to Example I. in Traverse 
Sailing, 



EXAMPLE VIIL 

Required the latitude in which the 
number of miles in a degree of longitude 
is =47.19443. 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

38° 00' = 47.28065 Next greater 
38 10 =47.17293 " less 



10 



.10772 

Given No. 47.19443 

Next less, 47.17293 



.02150 
10 



). 21500(2' nearly. 
21544 



38° 10'= 47.17293 Next less 
— 02 = + .02150 



58 08 = 47.19443 Given number. 



This refers to Example II. Part I. in 
Traverse Sailing. 



ON THE TABLE OF MILES IN A DEGREE OF LONGITUDE. 



21 



EXAMPLE IX. 

Requiretl the latitude in vrhich the 
number of miles in a degree of longitude 
is = 32.14623 

By the Table of IMiles in a degree of 
longitude, Sec, on page 9, 



[0 =32.09059 



less 



10 



.14734 



Given No. 32.14623 
Next less, 32.09059 



.05564 
10 



).55640(4^ nearly. 
58936 



57° 40^ = 32.09059 Next less 
_ 04 = + .05564 



57 36 



32,14623 Given number. 



This refers to Example II. Part II. in 

Traceme Sailing, 



EXAMPLE X. 

Required the latitude in which the 
number of miles in a degree of longitude 
is = 57.50771 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

16° 30^ = 57.52920 Next greater 
16 40 =57.47934 " less 



10 = .04986 



Given No. 57.50771 
Next less, 57.47934 





.02837 
10 




).28370(6^ nearly. 
29916 


16° 40^ 
— 06 


= 57.47934 Next less 

= + .02837 


16 34 


= 57.50771 Given number. 



EXAMPLE XL 

Required the latitude in which the 
number of miles in a degree of longitude 
is = 48.69213 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

35° 40^ = 48.74528 Next greater 
35 50 =48.64337 " less 



10 



.10191 

Given No. 48.69213 
Next less, 48.64337 



.04876 
10 



.48760(5^ nearly. 
50955 



35° 50^ 
— 05 



48.64337 Next less 
+ .04876 



35 45 = 48.69213 Given number.. 



This refers to Example III. Part I. in 
Traverse Sailing. 



EXAMPLE XIL 

Required the latitude in which the 
number of miles in a degree of longitude 
is = 57.78011 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

15° 30^ = 57.81787 Next greater 
15 40 =57.77092 " less 



10 = .04695 



Given No. 57.78011 
Next less, 57.77092 



.00919 
10 



).09190(2^ nearly. 
09390 



15° 40^= 57.77092 Next less 
— 02 = + .00919 



15 38 = 57.78011 Given number. 



This refers to Example II. Part III. in 
Traverse Sailing. 



This refers to Example III, Part II. in 

Traverse Sailing. 



00 



ON THE TABLE OF MILES IN A DEGREE OF LONGITUDE. 



EXAMPLE XIIL 



Required the latitude in which the 
number of miles in a degree of longitude 
is = 31.10492 

By the Table of Miles in a degree of 
longitude, &o., on page 9, 

58° 40^ = 31.20094 Next greater 
58 50 =31.05173 " less 



10 



.14921 





'50^ 
-08 


Given No. 
Next less, 

= 3L05173 
= + .11319 


31.16492 
31.05173 




.11319 
10 




)1.13190(8^ nearly. 
119368 


58^ 


Next less 


58 


42 


= 31.164D2 Given number. 



This refers to Example III. Part III. in 

Traverse Sailing. 



EXAMPLE XrV. 

Required the latitude in which the 
number of miles in a degree of longitude 
i8 = 38.97143 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

49° 20^ = 39.09941 Next greater 
49 30 =38.96685 " less 



10 = .13256 



Given No. 38.97143 
Next less, 38.96685 





30^ 
-00 


= 38.96685 
= + .00458 


.00458 
10 


49= 


).04580 
Next less 


49 


30 


= 38.97143 Given number. 



This refers to Example of Case III. Par- 
allel Sailing. 



EXAMPLE XV. 

Required tlie latitude in which the 
number of miles in a degree of longitude 
is = 43.54839 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

43° 20^ = 43.64242 Next greater 
43 30 =43.52240 " less 



10 



.12002 



Given No. 43.54839 
Next less, 43.52240 





30^ 
•02 


= 43.52240 Next less 
= + .02599 


.02599 
10 




).25990(2' 
24004 


43= 




43 


28 


= 43.54839 Given number. 



This refers to Question IV. for exercise 
in Parallel Sailing. 



EXAMPLE XVL 

Required the latitude in which the 
number of miles in a degree of longitude 
is = 8.67435 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

81° 40^ = 8.69592 Next greater 
81 50 =8.52320 " less 



10 = .17272 



Given No. 8.67435 
Next less, 8.52320 

.15115 
10 



)1.51150(9^ nearly. 
155448 



81° 50^ = 8.52320 Next less 
— 09 = + .15115 



81 41 



8.67435 Given number. 



This refers to Example of Case I. Mid^ 
die Latitude Sailing. 



ON THE TABLE OF MILES IN A DEGREE OF LONGITUDE. 



23 



EXAMPLE XVIL 

Required the latitude in which the 
numl)er of miles in a degree of longitude 
is = 38.03327 

Bj the Table of Miles in a degree of 
longitude, &c., on page 9, 



50° 30^ = 38.16474 Next greater 
50 40 =38.02982 " less 



10 



.13492 



Given No. 38.03327 
Next less, 38.02982 







39= 


10^ 

-08 


= 46.51872 
^ + .08593 






.00345 
10 


.08593 
10 








).03450(0^ 

== 38.02982 Next less 
= + .00345 


).85930(8^ nearly 
88080 


50^ 40^ 
— 00 


Next less 


50 40 


= 38.03327 Given number. 


39 


02 


= 46.60465 


Griven number. 









This refers to Example of Case II. Mid- 
dle Latitude Sailing. 



EXAMPLE XYIIL 

Required the latitude in which the 
number of miles in a degree of longitude 
is = 48.00000 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

Se'' 50^ = 48.02297 Next greater 
37 00 =47.91813 " less 



10 = .10484 

Given No. 48.00000 
Next less, 47.91813 





00^ 
-08 


= 47.91813 

= + .08187 


.08187 
10 






).81870(8^ 
83872 


nearly. 


37= 


Next less 
Given numl 




36 


52 


= 48.00000 


)er. 



This refers to Example of Case V. Mid- 
dle Latitude and Mercator's Sailing. 



EXAMPLE XIX, 

Required the latitude in which the 
numl)er of miles in a degree of longitude 
is = 46.60465 

By the Table of Miles in a degree of 
long'itude, &c., on page 9, 



39° 00^ 
39 10 



46.62882 Next greater 
46.51872 ** less 



10 = .11010 

Given No. 46.60465 
Next less, 46.51872 



This refers to Example of Case VII. 
Middle Latitude and Mercator's Sailing. 



EXAMPLE XX. 

Required the latitude in which the 
number of miles in a degree of longitude 
is = 59.69697 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 



5° 40^ 
5 50 


= 59.70671 Next greater 
= 59.68918 " less 


10 


= .01753 




Given No. 59.69697 
Next less, 59.68918 




.00779 
10 




).07790(4^ 
07012 



5° 50^ = 59.68918 Next less 
— 04 = + .00779 



5 46 =59.69697 Given number. 



This refers to Example of Case VIII., 
and Question X. for exercise, Middle 
Latitude Sailing. 



24 



ON THE TABLE OF MILES IN A DEGREE OF LONGITUDE. 



EXAMPLE XXL 

Required the latitude in which the 
number of miles in a degree of longitude 
is = 52.8544G 

Bj the Table of Miles in a degree of 
longitude, &c., on page 9, 

28° 10^ = 52.89465 Next greater 
28 20 =52.81207 " less 



10 = .08258 



Given No. 52.85446 
Next less, 52.81207 





20^ 
-05 


= 52.81207 
= + .04239 


Next less 
Given nur 


.04239 
10 




).42390(5 
41290 


28^ 


1100 


28 


15 


= 52.85446 


nber. 



This refers to Example of Case VIII. 
Middle Latitude Sailing, 



EXAMPLE XXIL 

Required the latitude in which the 
number of miles in a degree of longitude 
is = 32.74389 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

56° 50^ = 32.82455 Next greater 
57 00 =32.67835 " less 



10 



.14620 



Given No. 32.74389 
Next less, 32.67835 

.06554 
10 



).65540(4^ 
58480 



7060 



EXAMPLE XXin. 

Required the latitude in which the 
number of miles in a degree of longitude 
is = 53.23529 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

27° 20^ = 53.30098 Next greater 
27 30 =53.22062 " less 



10 = .08036 



Given No. 53.23529 
Next less, 53.22062 



.01467 
10 



.14670(2^ nearly. 
16072 



27° 30^= 53.22062 Next less 
— 02 = + .01467 



27 28 = 53.23529 Given number. 



57° 00^ = 32.67835 Next less 
— 04 = + .06554 



56 56 = 32.74389 Given number. 



This refers to Example of Case IX. 
Middle Latitude Sailing. 



This refers to Example of Case IX. 
Middle Latitude Sailing 



EXAMPLE XXrV. 

Required the latitude in which the 
number of miles in a degree of longitude 
is = 10.25782 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 



80° 00^ 
80 10 



10.41887 Next greater 
10.24696 " less 



10 = .17191 



Given No. 10.25782 
Next less, 10.24696 



.01086 
10 



).10860(K nearly. 
17191 



80° 10^= 10.24696 Next less 
— 01 = + .01086 



80 09 = 10.25782 Given number. 



This refers to Question I. for exercise 
in Middle Latitude Sailing. 



ON THE TABLE OF MILES IN A DEGREE OF LONGITUDE. 



25 



EXAMPLE XXV. 

Roquired the latitude in vrhich the 
number of miles iu a degree of longitude 
is = 10.65335 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 



79° 40' 
79 50 


= 10.76248 Next greater 
= 10.59073 ^' less 


arly. 


39° 10^ 
39 20 

10 

39° 20^ 
— 09 


= 46.51872 Next greater 
= 46.40819 " less 

= .11053 


10 


= .17175 

Given No. 10.65535 
Next less, 10.59073 


Given No. 46.51163 
Next less, 46.40819 

.10344 




.06462 
10 


10 




)1.03440(9^ 




).64620(4^ne 
68700 


99477 




3963 




= 10.59073 Next less 
= + .06462 




79° 50^ 
— 04 


= 46.40819 Next less 
= + .10344 




= 10.65535 Given number. 




79 46 


39 11 


= 46.51163 Given number. 









This refers to Question II, for exercise 
in Middle Latitude Sailing. 



EXAMPLE XXVL 

Required the latitude in vrhich the 
number of miles in a degree of longitude 
is =37.69058 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

51° 00^ = 37.75922 Next greater 
51 10 =37.62339 " less 



10 = .13583 

Given No. 37.69058 
Next less, 37.62339 





10^ 
05 


.06719 
10 






).67190(5^ ne 
67915 


arly 


51° 


= 37.62339 Next less 
= + .06719 




51 


05 


= 37.69058 Given number. 





This refers to Question Y. for exercise 
m Middle Latitude Sailing, and Question 
I. for exercise in Mercaior's Sailing. 



EXAMPLE XXVIL 

Required the latitude in which the 
number of miles iu a degree of longitude 
is = 46.51163 

By tlie Table of Miles in a degree of 
longitude, &c., on page 9, 



This refers to Question IX. for exercise 
in Middle Latitude Sailing and Question 
V. for exercise in Mercator's Sailing. 



EXAMPLE XXVIIL 

Required the latitude in which the 
number of miles in a degree of longitude 
is = 22.33885 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

68° 00^ = 22.47637 Next greater 
68 10 =22.31443 " less 



10 



.16194 

Given No. 22.33885 
Next less, 22.31443 



.02442 
10 



).24420(2^ nearly. 
32388 



68° 10^ = 22.31443 Next less 
— 02 = + .02442 



68 08 = 22.33885 Given number. 



Tills refers to Example of Compound 
Courses at the end of Middle Latitude 
and Mercaior's Sailing. 



26 



ox THE TABLE OF MILES IN A DEGREE OF LONGITUDE. 



EXAMPLE XXIX. 



Required the latitude in vrhich the 
number of miles in a degree of longitude 
is = 8.70430 



By the Table of Miles in a degree of 
longitude, &c., on page 9, 



81° 30^ = 8.86855 Next greater 
81 40 =8.69592 " less 





10 

40^ 
-00 




.17263 

8.69592 
+ .00838 


Given No. 
Next less, 

Next less 
Given nu 


8.70430 
8.69592 




.00838 
10 


8r 


).08380 


81 


40 


= 


8.70430 


mber 



This refers to Example of Case I. Mer- 
cator's Sailing. 



EXAMPLE XXX. 

Required the latitude in which the 
number of miles in a degree of longitude 
is = 37.68368 

By the Table of Miles in a degree of 
longitude, &c., on page 9, 

51° 00^ = 37.75922 Next greater 
51 10 =37.62339 " less 



10 = .13583 



Given No. 37.683G8 
Next less, 37.62339 

.06029 
10 



). 60290(4^ 
54332 



5958 



51° 10^ = 37.62339 Next less 
— 04 = + .06029 



51 06 = 37.68368 Given number. 



This refers to Example of Case II. 3Ier- 
caior's Sailing. 



EXAMPLE XXXL 

Required the latitude in which the number of miles in a degree of longitude 
is = 25.98765. 

By the Table of Miles in a degree of longitude, &c., on page 9. 



64° 20^ = 25.98820 Next greater 
64 30 =25.83065 " less 



10 



.15755 



Given No. 25.98765 
Next less, 25.83065 

.15700 
10 



64° 30^ = 25.83065 Next less 
— 10 =^.15700 



64 20 = 25.98765 Given number. 



)1.57000 (9^9 -f 
141795 nearly 10 



152050 
141795 



10255 



This refers to Question X. for exercise, Middle Latitude Sailing. 



;:■ 



PLANE SAILING. 



Plane Sailing is that in -which the Earth and Sea are considered as constituting 
an extended plane, the meridians as being straight lines all parallel to each other, 
and the parallels of latitude as straight lines at right angles to the meridians. The 
length of a degree on the meridian and parallels of latitude is everywhere equal to 
what it is on the Equator, viz. 60 miles. In this sailing there are four principal 
parts, viz. the Course, Distance, Difference of Latitude, and Departure, any two of 
which being given, the others can be calculated by common arithmetic. 

The Course is the angle which a ship's track or path makes with the meridian, 
and is expressed either in points or degrees. Thus, when a ship sails N.N.E., her 
Course is 2 points or 22° 30^ 

The Complement of the Course is the difference between the Course and 8 points 
or 90°. Thus, the Course is N.N.E., 2 points or 22° 30^; therefore, the Complement 
of the Course is 6 points or 67° 30^. 

The Distance is the number of miles, leagues, &c., between any two places reck- 
oned on the Rhumb line of the course ; or it is the length that a ship has sailed on a 
direct course in a given time. 

The Difference of Latitude is the distance which a ship has made north or 
south of the parallel of the place sailed from, and is reckoned on a meridian. 

The Departure is the distance which a ship has made east or west of the meri- 
dian of the place sailed from, and is reckoned on a parallel of latitude. 

Hence, it is evident that if a ship sails due north or due south, she sails on a me- 
ridian, makes no departure, and her distance and difference of latitude are the same. 
If a ship sails due east or due west, she sails on a parallel of latitude, makes no dif- 
ference of latitude, and her distance and departure are the same. But when a ship 
sails in any other direction, she makes both difference of latitude and departure, 
and these, with the distance, form a right-angled triangle, the hypothenuse of which 
is the distance sailed, the perpendicular is the difference of latitude, the base is the 
departure, the angle opposite to the base is the course, and that opposite to the per- 
pendicular is the complement of the course ; hence, any two of these parts being 
given, the rest may be found by common arithmetic, as shown in the following 
pages. 

On a four-point course the ship makes as much difference of latitude as she does 
departure ; on a course less than four points she makes more difference of latitude 
than she does departure ; but on a course greater than four points she makes more 
departure than she does difference of latitude. 

The Horizon is that great and visible circle which is seen by an observer as if 
described around him, constituting the boundaries of his sight and apparently 
touching the heavens. It is distinguished by the terms visible, sensible, and rational 
or true. 

The visible horizon is that which is seen while the eye is elevated above the sur- 
face ; and the sensible is that which is seen when the eye is on a level with the sea. 
The depression of the former below the latter is caused by the elevation of the eye 

(27) 



28 PLANE SAILING. 

of the observer above the water, and is called the Dip of the visible horizon. The 
rational or true horizon is the circle in -which a plane parallel to the plane of the 
sensible horizon, and passing through the centre of the Earth, cuts the celestial 
concave. 

The Terrestrial Equator is a great circle supposed to be described around the 
Earth at the distance of 90° from the poles, dividing the Globe into tvro equal 
parts, the part to the northward of the Equator being called the northern hemisphere, 
ajid that to the southward the southern hemisphere. 

The Celestial Equator, commonly called the Equinoctial, is an imaginary circle 
described in the heavens, corresponding to the terrestrial Equator. 

The Terrestrial Poles, or poles of the Earth, are two points on the sur- 
face of the Globe, diametrically opposite to each other and distant from the 
Equator 90° ; the northernmost is called the north pole and its opposite the south 
pole. A straight line drawn through the Earth from one of these points to the 
other is called the axis of the Earth, as around it our planet performs its diurnal 
revolution from west to east, causing the apparent movement of the heavenly bodies 
from east to west. 

The Celestial Poles, or poles of the World, are two imaginary points in the hea- 
vens directly opposite to the poles of the Earth ; and a straight line extending from 
one of these points to the other, and passing through the poles of the Earth, is 
called the axis of the World, around which all the celestial bodies appear to perform 
their revolution from east to west. 

Terrestrial Meridians are great circles supposed to be drawn on the surface of 
the Globe, at right angles to the equator and meeting at the poles. They are some- 
times called circles of longitude ; and all places situated on the same meridian have 
the same longitude. 

A Great Circle of the sphere is one whose plane passes through the centre of 
the sphere, such as the Equator, the Meridians, &c. 

A Small Circle of the sphere is one whose plane does not pass through the cen- 
tre of the sphere, such as the Polar Circles, the Tropics, Parallels of Latitude, &c. 

The Latitude of any place is its distance north or south of the Equator, and is 
measured by that portion of a terrestrial meridian which is intercepted between that 
place and the Equator. 

Parallels of Latitude are small circles supposed to be drawn parallel to the 
Equator; and all places situated on the same parallel are said to have the same 
latitude. 

As latitude is counted from the Equator towards the poles on a terrestrial meri- 
dian, North and South, the Difference of Latitude between two places, both North 
or both South, is found by suhtraciiny the less Latitude from the greater ; but if one 
Latitude be North and the other South, the Difference of Latitude is found by add- 
ing both Latitudes together. 

In North latitude sailing to the Northward, or in South latitude sailing to the 
Southward, the Difference of Latitude must be added to the Latitude left to get the 
Latitude in, because the ship sails farther away from the Equator. 

But in North latitude sailing to the Southward, or in South latitude sailing to the 
Northward, the Difference of Latitude must be subtracted from the Latitude left to 
get the Latitude in, because the ship sails nearer to the Equator. 

When the Latitude decreases and the Difference of Latitude is greater than the 
Latitude left, the Latitude left must be subtracted from the Difference of Latitude, to 
get the Latitude in, which will be of a different name from the Latitude left, because 
the ship in this case has crossed the Equator. 



PLANE SAILING. 29 

CASE I. 
Given the Course and Distance to find the Difference of Latitude and Departure. 

RULE. 

Isf. Tojind the Difference of Latitude. 
Multiply the Distance in miles by the number of miles in a degree of longitude 
in latitude equal to the Course, and^divide the product by 60; the quotient will be 
the number of miles in the Difference of Latitude. 

2c?. Tojind the Departure. 
Multiply the Distance in Miles by the number of miles in a degree of longitude 
in latitude equal to the Complement of the Course, and divide the product by 60; 
the quotient will be the number of miles in the Departure. 

EXAMPLE. 
A ship from ^he latitude of 49° 57^ North sails S.W. by W. 244 miles ; required 
the latitude she is in and her Departure from the meridian. 

By the Table of Points opposite to the Compass, on page 10, 
Course S.W. by W. ... 5 points = 56° 15' gives 33.33418 the mimber of miles in a degree of longitude 

in latitude eqiial to it. 
Complement of the Course 3 " = 33 45 " 49.88810 the nuDiber of miles in a degree of longitude 

in latitude equal to it. 

\st. Tojind the Difference of Latitude. 

244 X 33.33418 ^ -^35559 nearly, the number of miles in the Difference of Latitude = — = 2^ IG' S. 
60 60 

Which subtracted from the Latitude left 49 57 N. 

Gives the Latitude the ship is in . . . 47 41 X. 

Answer. 

2d. To find the Departure. 

— — = 202.878 the number of miles in the Departure = 202.9 Answer. 

60 

CASE IL 

Given the Course and Difference of Latitude to find the Distance and Departure. 

RULE. 
Is^. To find the Distance. 
Multiply the Difference of Latitude in miles by 60, and di^ade the product by the 
number of miles in a degree of longitude in latitude equal to the Course ; the quo- 
tient will be the number of miles in the Distance. 

2c?. To find the Departure. 
Multiply the Distance in miles by the number of miles in a degree of longitude 
in latitude equal to the Complement of the Course, and divide the product by 60 ; 
the quotient will be the number of miles in the Departure. 

EXAMPLE. 
A ship runs S.E. by E. from 1° 45^ North latitude, and then by observation is in 
0° ZV South latitude ; required the Distance and Departure. 

By the Table of Points opposite to the Compass, on page 10, 

Course S.E. by E 5 points = 56° 15' give 33.33418 the number of miles in a degree of longitude in 

latitude equal to it. 
Complement of the Course 3 " = 33 45 " 49.88810 the number of miles in a degree of longitude in 

latitude equal to it. 
Latitude left . . . .1° 45' N. 
in .... 31 S. 



Difference of Latitude 2 16 = 136 miles in the Difference of Latitude. 

\st. To find the Distance. 

136 X 60 

-— — — — = 244.794 the number of miles in the Distance = 244.8 Answer. 

33.33418 ; 

2c?. To find the Departure. 

244.8 X 49.88810 

— = 203.542 the number of miles in the Departure = 203.5 Answer. 



30 PLANE SAILING. 

CASE IIL 

Given the Course and Departure to find tlie Distance and Difference of Latitude. 

RULE. 
IsL To find the Distance. 
Multiply the Departure in miles Ly 60, and divide the product by the number of 
miles in a degree of longitude in latitude equal to the Complement of the Course ; 
the quotient will be the number of miles in the Distance. 

2d. To find the Difference of Latitude. 
Multiply the Distance in miles by the number of miles in a degree of longitude 
in latitude equal to the Course, and divide the product by 60 ; the quotient will be 
the number of miles in the Difference of Latitude. 

EXAMPLE. 
If a ship sails N.E. by E. f E. from a port in 3° 15^ South latitude until she 
departs from her first meridian 203 miles (the Departure), required the Distance 
sailed and the latitude she is in. 

By the Table of Points opposite to the Compass, on page 10, 

Course N.E. by E. :J E, . , 5f points = 64° 41^' gives 25.65329 the number of miles in a degree of longitude 

in latitude equal to it. 

Complement of the Course 2i " = 25 18| " 54.23930 the number of miles in a degree of longitude 

in latitude equal to it. 

l5^. To find the Distance. 

203 X 60 

- , . ^„„„^ = 224.560 the number of miles in the Distance sailed = 224.6 Answer. 

54.2:3930 

2d. To find the Difference of Latitude. 

224.6 X 25.65329 96 

= 96.029 nearly, the number of miles in the Difference of Latitude = — = 1° 36' N. 

"Which subtracted from the Latitude left 3 15 S. 



Gives the Latitude the ship is in ... 1 39 S. 

Answer. 

CASE IV. 
Given the Distance and Difference of Latitude to find the Course and Departure. 

RULE. 
1st. To find the Course. 
Multiply the Difference of Latitude in miles by 60, and divide the product by the 
Distance in miles ; the quotient will be the number of miles in a degree of longi- 
tude in latitude equal to the Course. 

2d. To find the Departure."^ 
Multiply the Distance in miles by the number of miles in a degree of longitude 
in latitude equal to the Complement of the Course, and divide the product by 60 ; 
the quotient will be the number of miles in the Departure. 

EXAMPLE. 

Suppose a ship sails 244 miles between the South and East from a port in 2° 52'' 
South latitude, and then by observation is in 5° 08^ South latitude, what Course has 
she steered and what Departure has she made ? 

Latitude left .... 2"= 52' S. 
in .... 5 08 S. 



Difference of Latitude 2 16 = 136 miles in the Difference of Latitude. 

1st. To find the Course. 

136 X 60 

• — r— — == 33.44262 the number of miles in a degree of longitude in latitude equal to the Course, viz. 
^^ S. 56° 08' E., as per Example 1., page 19. 

The Complement of the Course, viz. 33° 52', gives 49.82016, the number of miles in 8 
degree of longitude in latitude equal to it, as per Example I., page 13. 

2d. To find the Departure. 

244 X 49.82016 

— ^ = 202.602 nearly, the number of miles in the Departure = 202.6 Ans^wer. 

60 

* The Departure may be found otherwise, thus : — 
l/(Distance)2 — (Dif. of Lat.)2 = |/(244)2 — (136)2 = i/59536 — 18496 = i/41040 202.583 the number 

of miles in the Departure -= 202.6 Answer. 



PLANE SAILING. 31 

CASE V. 

Given the Distance and Departure to find the Course and Difference of Latitude. 

RULE. 

Isi. Tojind the Course, 

Multiply the Departure in miles by 60, and divide the product by the Distance in 
miles ; the quotient vrill be the number of miles in a degree of longitude in latitude 
equal to the Complement of the Course. Subtract the Complement from 90°, and 
the remainder will be the Course. 

2d. Tojind the Difference of Latitude* 
Multiply the Distance in miles by the number of miles in a degree of longitude 
in latitude equal to the Course, and divide the product by 60 ; the quotient will be 
the number of miles in the Difference of Latitude. 

EXAMPLE. 

Suppose a ship sails 244 miles between the North and West, from the latitude of 
32° 2y North, until her Departure is 203 miles ; what Course has she steered, and 
what latitude is she in ? 

\st. Tojind the Course. 

_"- r= 49.91S03 the number of miles in a degree of longitude in latitude equal to the Complement of the 

2^ Course, viz. 33° 42', as per Example II., page 19. 

The Course, viz. N. 56° 18' W., gives 3:i.'290f^3, the number of miles in a degree of lon- 
gitude in latitude equal to it, as per Example 11., page 13. 

2d. Tojind the Difference of Latitude. 

24^ y ^3 *^9063 * 13t 

■ — = 135.382 nearly, the number of miles in the Difference of Latitude = — '- = 2° 15' N. 

60 60 

Which added to the Latitude left 32 25 N. 



Gives the Latitude the ship is in 34 40 N, 
Ans-sver. 

CASE VL 
Given the Difference of Latitude and Departure to find the Distance and Course. 

RULE. 

\si. To find the Distance. 

Add the square of the Difference of Latitude in miles to the square of the De- 
parture in miles, and from the sum of these two squares extract the square root, 
which will be the number of miles in the Distance. 

2d. To find the Course. 
Multiply the Difference of Latitude in miles by 60, and divide the product by the 
Distance in miles ; the quotient will be the number of miles in a degree of longitude 



in latitude equal to the Course. 



EXAMPLE. 



A ship sails between the North and West till her Difference of Latitude is 136 
miles and her Departure is 203 miles; required her Distance and Course. 

1.9^. To find the Distance. 



y^(Dif. of Lat.)2 + (Departure)^ = ■j/(136)2 + (203)2 = |/l8496 + 41209 = 1/ 59705 = 244.346 the num- 
ber of miles in the Distance. Answer. 

2d. To find the Course. 

136 X 60 
.,,1 o.R = 33.39527 nearly, the number of miles in a degree of longitude in latitude equal to the Course, 
-■"•'**^ viz. N. 56° II' W., as per Example IlL, page 19. Answer. 



• The Difference of Latitude may be found otlierwise, tlm^ : — 



|/(Distauce)2 - (Depaxture)^ = -|/' (244)2 _ (^203)2 = -j/ 59536 - 41209 = -j/ 18327 = 135.377 the number 

of miles in the Difference of Latitude = 135.4 Answer. 



82 PLANE SAILING. 

QUESTIONS 

To exercise the Learner in the foregoing Rules. 

QUESTION L 

A ship in 2° 10^ South latitude sails N. by E. 89 leagues (= 267 miles) ; what 
Latitude is she in, and what is her Departure from the meridian ? 

By the Table of Points opposite to the Compass, on page 10, 

Course N. by E 1 point = 11° 15' gives 58.84702 the number of miles in a degree of longitude in 

latitude equal to it. 
Complement of the Course 7 points = 78 45 " 11.70539 the number of miles in a degree of longitude in 

latitude equal to it. 

BY CASE I. 

1st. To -find the Difference of Latitude. 

267 X 58.84702 ^ ^ ^gg 

— = 261.871 the number of miles in the Difference of Latitude = -— = 4° 22' N. 

60 WJ 

From which suUract (because the ship has crossed the Equator) the Latitude left 2 10 S. 



And the remainder will be the Latitude in 2 12 N. 

Answer. 

2d. Tofnd the Departure. 

267 X 11.70539 52.089 

• — • = 52.089 the number of miles in the Departure = — - — = 17.363 leagues. Answer. 

QUESTION IL 

A ship sails S.S.W. from a port in 4P 30^ North latitude, and then by obserTatioa 
is in 36° 57 •^ North latitude ; required the Distance run and her Departure from the 
meridian. 

By the Table of Points opposite to the Compass, on page 10, 

Course S.S.W 2 points = 22° 30' gives 55.43268 the number of miles in a degree of longitude 

in latitude equal to it. 

Complement of the Course 6 " = 67 30 " 22.96101 the number of miles in a degree of longitude 

in latitude equal to it. 



BY CASE II. 



Latitude left . . . . 41° 30' N. 
" in .... 36 67 N. 



Difference of Latitude 4 33 = 273 miles in the Difference of Latitude. 

1st. To find the Distance. 

273 X 60 295.5 

.. ^ = 295.494 the number of miles in the Distance = — — - = 98.5 leagues. Answer. 
5o.4o2do o 

2d. To find the Departure. 

295.5X22.96101 „„ „„ , 11.3.1 „„ „ , 

— = 113.083 the number of miles m the Departure = — ;: — = 37.7 leagues. Answer, 

60 3 

QUESTION IIL 

A ship sails S.S.W.^ "W. from a port in 2° 30^ South latitude, until her Departure 
is 59 leagues (= 177 miles) ; required the Distance run and Latitude in. 

By the Table of Points opposite to the Compass, on page 10, 

Course S.S.W. ^ W. . . . 2^ points = 28° 07^' gives 52.91519 the number of miles in a degree of longitude 

in latitude equal to it. 

Complement of the Course 5i " = 61 52| " 28.28376 the number of miles in a degree of longitude 

in latitude equal to it. 

BY CASE III. 

1st. To find the Distance. 

177 X 60 375 5 

• J^ - = 375.470 the number of miles in the Distance = — -^ = 125.2 leagues. Answer. 



2d. To find the Difference of Latitude. 

:he Difference of Latil 
Which added to the Latitude left 2 30 S. 



— ■ ' = 331.161 the number of miles in the Difference of Latitude = ^ = 5° 31' S. 



Gives the Latitude the ship is in 8 01 S. 
Answer. 



PLANE SAILING. 33 

QUESTION IV. 

If a ship sails 360 miles South-Westward from 21° 59^ South latitude, until by 
observation she be in 24° 49^ South latitude, what is her Course and Departure from 
the meridian ? 



BY CASE IT. 



Latitude in .... 24° 49' S, 
left . . . . 2i 59 S. 



Difference of Latitude 2 50 = 170 miles in the Difference of Latitude. 

Isi. To find the Course. 

— = 28.33333 the number of miles in a degree of longitude in latitude equal to the Course, viz. 

36*^ S. 61° 49' W., as per Example IV., page 19. 

The Complement of tlie Course, viz. 28° 11', gives 52.88639, the number of miles in a 
degree of longitude in latitude equal to it, as per Example III., page 14. 

2d. To find the Departure. 

360 y 62 88639 

— ^ = 317.318 the number of miles in the Departure = 317.3 Answer. 



QUESTION V. 

Suppose a ship sails 354 miles North-Eastward from 2° 09'' South latitude until 
her Departure be 160 miles, what is her Course and Latitude in? 

BY CASE V. 

\st. To find the Course. 

— — - — = 25.42373 the number of miles in a degree of longitude in latitude equal to the Complement of th* 
*^ Course, viz. 64° 56', as per Example V., page 20. 

The Course, viz. N. 25° 04' E., gives 54.34885, the number of miles in a degree of lon- 
gitude in latitude equal to it, as per Example IV., page 14. 

2d. To find the Difference of Latitude. 

354 X 54.34885 ^ ggo.gss the number of miles in the Difference of Latitude = ^ = 5° 21' N. 
60 60 

From which subtract (because the ship has crossed the Equator) the Latitude left 2 09 S. 



And the remainder will be the Latitude in 3 12 N. 

Answer^ 



QUESTION VL 

Sailing between the North and West from a port in 1° 5^'' South latitude, and 
then arriving at another port in 4° 08^ North latitude, which is 209 miles (the De- 
parture) to the Westward of the first port, required the Distance and Course from 
the first port to the second. 



Latitude left . . 
« in . . 


. . 10 59'S. 
. . 4 08 N. 


Difference of Latitude 6 


07 = 367 miles in the Di 






BY CASE VI. 




1st. 


To find the Distance. 



■j/(367)2 + (209)2 = 1/134689 + 43681 = y'llSSlO = 422.339 the number of miles in the Distance = 

422 339 

— ~ — = 140.779 = 140.8 leagues. Answer. 



2d. To find the Course. 

i2.13821 the number of n 
N. 29° 40' W., or ] 

3 



' ioooQQ "= 52.13821 the number of miles in a degree of longitude in latitude equal to the Course, viz. 
^^-"^ N. 29° 40' W., or N.N.W. | W. nearly, as per Example VI., page 20. Answer. 



34 PLANE SAILING. 

QUESTION Vn. 

_ Four dajs ago we were in latitude 3° 25^ South, and have since that time sailed in a 
direct Course N.W. by N. at the rate of 8 miles an hour (making the Distance 768 
miles) ; required our present Latitude and Departure. 

By the Table of Points opposite to the Compass, on page 10, 

Course N.W. by N. ... 3 points = 33° 45' gives 49.88810 the number of miles in a degree of longitude in 

latitude equal to it. 
Complement of the Course 5 " =56 15 " 33.33418 the number of miles in a degree of longitude in 

latitude equal to it. 

BY CASE I. 

1st To find the Difference of Latitude. 

768 "i^ 49 88810 fi?Q 

—^ = 638.568 the number of miles in the Difference of Latitude = — = 10° 39' N. 

60 60 

From which subtract (because the ship has crossed the Equator) the Latitude left 3 25 S. 



And the remainder will be the Latitude in 7 14 N. 

Answer. 

2d. To find the Departure. 

768 X 33.33418 

■ — = 426.678 the number of miles in the Departure = 426.7 Answer. 

QUESTION VIIL 

A ship in the latitude 3° 52^ South is bound to a port bearing N.W. by W. J W., 
in the latitude 4° 30'' North ; what is the ship's Distance from the port, and how far 
does it lie to the Westward (the Departure) ? 

By the Table of Points opposite to the Compass, on page 10, 

Course N.W. by W. i W. . 5i points = 61^ 52i-' gives 28.28376 the number of miles in a degree of longitude 

in latitude equal to it. 
Complement of the Course 2^ " = 28 871 " 52.91519 the number of miles in a degree of longitude 

ia latitude equal to it. 
Latitude the ship is in 3° 52' S. 
" bound to . . 4 30 N, 



Difference of Latitude 8 22 = 502 miles in the Difference of Latitude. 
BY CASE II. 

1^. To find the Distance. 

— ^zrr: = 1064.918 the number of miles in the Distance = 1065. Answer. 
28.28376 

2d. To find the Departure. 

1064 918 V 52 91519 

: : = 939.172 the number of miles in the Departure = 939.2 Answer. 



QUESTION IX. 

A ship from the latitude of 48° 17^ North sails S. W. by S. until she has depressed 
the North Pole 2° (made a Difference of Latitude of 120 miles) ; what direct Dis- 
tance has she sailed, and how many miles has she got to the Westward (the De- 
parture) ? 

By the Table of Points opposite to the Compass, on page 10, 

Course S.W. by S. ... 3 points = 33° 45' gives 49.88810 the number of miles in a degree of longitude in 

latitude equal to it. 
Complement of the Course 5 " «= 56 15 " 33.33418 the number of miles in a degree of longitude in 

latitude equal to it. 

BY CASE II. 



1st. To find the Distance. 

the number of miles in the Distan 

2d. To find the Departure. 



■ — — = 144.323 the number of miles in the Distance = 144.3 Answer. 

49.88810 



141 3''3 V 33 33418 

— '' "' =. 80.181 the number of miles in the Departure = 80,2 Ar.swer. 

60 



TRAVERSE SAILING. 



"When a vessel is obliged to sail on different courses, from contrary winds or other 
causes, the irregular track which she makes is called a Traverse ; and the method 
of finding out a single course and distance which would take her direct from the 
place of her Departure to where she has arrived is called Resolving a Traverse. 

A Traverse is resolved by finding the Difference of Latitude and Departure cor- 
responding to each Course and Distance sailed, as in Case I. Plane Sailing, and 
entering them in a table called a Traverse Table, the proper practical form of which 
is given in the following Examples. Care must )3e taken, when the ship sails to the 
Northward, to enter the Difference of Latitude in the column marked N., but 
when the ship sails to the Southward, to enter the Difference of Latitude in the 
column marked S. In like manner, when the ship sails to the Eastward care must 
be taken to enter the Departure in the column marked E., but when the ship sails to 
the AVestward to enter the Departure in the column marked W. 



If the ship sails due North or due South, she makes no Departure, and there- 
fore the whole Distance sailed will be Difference of Latitude, and must be placed in 
the column marked N. if she sailed to the North, but in the column marked S. if she 
sailed to the South. Again, if the ship sails due East or due West, she makes no 
Difference of Latitude, and therefore the whole Distance sailed will be Departure, 
and must be placed in the column marked E,, if she sailed to the East, but in the 
column marked W. if she sailed to the West.. 

When a ship is bound from one place to another, and is about losing sight of the 
land, the officers observe its bearing per Compass ((Supposed to be true), and estimate 
its distance from the ship. This is called taking the ship's Departure, and is always 
reversed and put down in the day's work as the first Course and Distance sailed. 
Some prominent point, island, lighthouse, &c., the latitude and longitude of which 
are known, must be selected for this purpose, and the ship's reckoning is said to 
begin at that place. 

RULE. 

Part I. The direct Distance and Course from one port or place to another, the 
Difference of Latitude and Departure being given, can be calculated by Case VI. 
Plane Sailing. 

Part II. Reverse the bearing of the land or Departure and consider it the first- 
Course onwhich the ship sailed the estimated Distance she was from it, and calcu- 
late the Difference of Latitude and Departure corresponding to said Course and Dis- 
tance by Case I. Plane Sailing. Calculate also the Difference of Latitude and De- 
parture for each Course and Distance sailed, in the same manner. 

Make a Traverse Table consisting of six columns, and head them Course, Distance, 
N. S. E. W., beginning at the left side, and write down the several Courses, Dis- 
tances, Differences of Latitude, and Departures, in their respective columns. Add 
up the Northings, Southings, Eastings, and Westings separately, and set down the 
sum of each at the bottom of its proper column. If the Northings be greater than 
the Southings, subtract the Southings from the Northings, and the remainder will 
be the Difference of Latitude made good, to the Northward ; but if the Southings 
be greater than the Northings, subtract the Northings from the Southings, and the 
remainder will be the Difference of Latitude made good to the Southward. Again, 
if the Eastings be greater than the Westings, subtract the Westings from the East- 
ings, and the remainder will be the Departure made good to the Eastward ; but if 

(35) 



36 TRAVERSE SAILING. 

the Westings be greater than the Eastings, subtract the Eastings from the Westings, 
and the remainder will be the Departure made good to the Westward. 

With the Difference of Latitude made good and the Departure made good, calcu- 
late the Distance made good and the Course made good, which would have taken the 
ship direct from the place of her departure to where she has arrived, by Case VI. 
of rlane Sailing. 

Part III. From the whole Difference of Latitude in miles, to be made, subtract 
tlie Difference of Latitude in miles made good, and the remainder will be the Differ- 
ence of Latitude in miles yet to make. And from the whole Departure in miles, to 
be made, subtract the Departure in miles made good, and the remainder will be the 
Departure in miles yet to make. With the Difference of Latitude in miles yet to 
make and the Departure in miles yet to make, the direct Distance in miles yet to 
make and the Course yet to make can be calculated by Case VI. Plane Sailing, 

EXAMPLE L 

Suppose a ship takes her departure from Block Island, the middle of it bearing 
N.N.W., distant by estimation about 15 miles (this is to be reversed and considered 
the Ist Course), and in the Latitude of 41° 10^ North, sails, 2d, S.E. 34 miles ; 
3d, W. by S. 16 miles ; 4th, W.N.W. 39 miles ; and 5th, S. by E. 40 miles : required 
her Difference of Latitude made good, the Latitude she is in, her Distance from 
Block Island and the bearing of Block Island from the ship. (Part I. of the Rule 
is not required in this Example.) 

Part II. 

To find the Distance and Course made good hy the Ship. 

By the Table of Points opposite to the Compass, on page 10, 

let Course S.S.E 2 points = 22° 30' gives 55.43268 the nnmber of miles in a degree of longitude 

in latitude equal to it. 
Complement of the Course 6 " =67 30 " 22.96101 the number of miles in a degree of longitude 
* in latitude equal to it. 

BY CASE I. PLANE SAILING. 

1st. To find the Difference of Latitude on the \st Course. 

15 X 55.43268 

-~- = 13.85817 the number of miles in the Difference of Latitude = 13.9 Answer. 

2d. To find the Departure on the \st Course. 

15 X 22.96101 



60 



= 5.74025 the number of miles in the Departure = 5.7 Answer. 



By the Table of Points opposite to the Compass, on page 10, 

2d Course S.E. ..... 4 points = 45° gives 42.42637 the number of miles in a degree of longitude in 

latitude equal to it. 
Complement of the Course 4 " =45 " 42.42637 the number of miles in a degree of longitude in 

latitude equal to it. 

BY CASE I. PLANE SAILING. 

\st. To find the Difference of Latitude on the 2d Course. 

— — — '- = 24.04161 the number of miles in the Difference of Latitude = 24 Answer. 

60 

2d. To find the Departure on the 2d Course. 

34 X 42.42637 

'- = 24.04161 the number of miles in the Departure = 24 Answer. 



By the Table of Points opposite to the Compass, on page 10, 

3d Course W, by S. . . . 7 points = 78° 45' gives 11.70539 the number of miles in a degree of longitude in 

latitude equal to it. 
Complement of the Course 1 point = 11 15 " 58.84702 the number of miles in a degree of longitude in 

latitude equal to it. 

BY CASE I. PLANE SAILING. 

\st. To find the Difference of Latitude on the M Course. 

16 X 11.70539 

-^ = 3.12144 the number of miles in the Difference of Latitude = 3.1 Answer. 

60 

2d. To find the Departure on the 2>d Course. 

• — ^ ■= 15.69254 the number of miles in the Departure = 15.7 Answer. 



TRAVERSE SAILING. 



87 



By the Table of Points opposite to the Compass, on page 10, 

4th Course W.N.W. ... 6 points = 07° 30' gives 2'2.96101 the nnnilier of miles in a degree of longitude in 

latitude equal to it. 
Complement of the Course 2 " =22 30 " 55.43268 the number of miles in a degree of longitude in 

latitude equal to it. 



BY CASE I. PLANE SAILING. 

l^^. To find the Difference of Latitude on the 4th Course. 

39 X 22.96101 



60 



1-1.92466 the number of miles in the Difference of Latitude = 14.9 Answer. 



2(7. To find the Departure on the 4ih Course. 

• — -^ — "^ = 36.03124 the number of miles in the Departure = 3( 



Answer. 



By the Table of Points opposite to the Compass, on page 10, 

5th Course S. by E. . . . 1 point = 11° 15' gives 58.84702 the number of miles in a degree of longitude in 

latitude equal to it. 
Complement of the Course 7 points = 78 45 " 11.70539 the number of miles in a degree of longitude in 

latitude equal to it. 



BY CASE I. PLANE SAILING. 

1st. To find the Difference of Latitude on the 5ih Course. 



40 X 58.84702 
60 



39.23135 the number of miles in the Difference of Latitude ■■ 



2d. To find the Departure on the 5th Course. 

40 X 11 70539 

- o^orr, ^jjg number of miles in the Departure = 7.8 Answer. 



Having calculated all the Diflferences of Latitude and Departures corresponding 
to the several Courses and Distances sailed, arrange them, according to the Eule, in 
the form of a 

TRAVERSE TABLE. 



Course. 


Distance. 


Difference 

N. 


OF Latitude. 

S. 


Depai 
E. 


ITURE. 

W. 


S.S.E. 

S.E. 

W. by S. 

W.N.W. 

S. by E. 


15 
34 
16 
39 
40 


14.9 


13.9 

24.0 

3.1 

39.2 


5.7 
24.0 

7.8 


15.7 
36.0 


Differs 


14.9 


80.2 
14.9 


37.5 


51.7 
37.5 


Remair 
mce of Latitu( 


ider 65.3 

le made good. 


Remaic 
Departu 


der 14.2 

re made good. 

1 



BY CASE VI. PLANE SAILING. 

1st. To find the Distance made good. 



65.3X60 



|/(65.3)2 + (14.2)2 = |/4264.09 + 201.64 = -j/4465.73 = 66.826 ^^^ number of miles in the Distance made 

good = 66.8 Answer. 

2d. To find the Course made good. 

= 58.62987 the number of miles in a degree of longitude in latitude equal to the Course made 
good, viz. S. 12° 16' W., because the DiiTerence of Latitude made good, is to the South- 
ward, and the Departure made good, is to the Westward, as per Example VII., page 20. 
Answer. Therefore, Block Island beai-s from the ship on the point of the compass di- 
rectly opposite, viz. N. 12"=' 16' E. 

The Latitude left (Block Island) ... 41° 10' N. 
Dif. of Latitude made good, 65.3 miles = 1 05 S. 



The Latitude the ship is in 40 05 N. 

Part III. of the Rule is not required in this Example. 



38 TRAVERSE SAILING. 

EXAMPLE XL 

A ship from Mount Dcsort Rock, in the latitude of 43° 50^ N., sails for Cape Cod, 
in the latitude of 42° 03^ N., its Departure from the meridian of Muunt Desert Hock 
being supposed to be 84 miles to the Westward. She takes her departure from 
Mount Desert Kock bearing due North, distant by estimation 10 miles, Ijut by reason 
of contrary winds she is obliged to sail on the following Courses, viz. (1st, South 
10 miles, the Departure reversed); 2d, W.S.W. 25 miles; 3d, S.W. 30 miles; and 
4th, West 20 miles : required, 1st, the direct Distance and Course from Mount De- 
sert Rock to Cape Cod ; 2d, the Distance and Course made good by the ship, from 
Mount Desert Rock to where she has arrived ; and 3d, the direct Distance and Course 
from where she now is, to Cape Cod, the place bound to. 

Bemark.—The latitude of Mount Desert Rock lighthouse is 43° 58^ N. ; the lati- 
tude of Cape Cod (Highland) light is 42° 02^ N. The meridian of Cape Cod is only 
83 miles to the Westward of the meridian of Mount Desert Rock. But as we want 
to work out all the Examples in Bowditch by our method, we must use his figures, 
although knowing them to be not strictly correct. 

Latitude of Mt. Desert Rock 43° 50' N. 
Cape Cod light . 42 03 N. 



Difference of Latitude . . 1 47 =107 miles. The Departure is 84 miles. 

Part I. 
To find the direct Distance and Course from the place left, to the port hound to. 

BY CASE VI. PLANE SAILING. 

\st. To find the Distance. 



-j/(107)a + (84)2 = i/ll449 + 7056 = |/lS505 = 136.033 the number of miles in the direct Distance from 

Mt. Desert Rock to Cape Cod. 

2d. To find the Course. 

107 X 60 ^ 

, oa (Yio = 47.19443 the number of miles in a degree of longitude in latitude equal to the direct Course, 
ido.u<w yj2. S. 38° 08' W., because the Difference of Latitude, is to the Southward, and the Depart- 

ure to the Westward, as per Example VIII., page 20. Answer. 

Therefore, Mount Desert Rock bears from Cape Cod, on the point of the compass di- 
rectly opposite, viz. N. 38° 08' E. 

Part II. 
To find the Distance and Course made good hy the ship, from Mount Desert Rock to 

where she has arrived. 

The bearing of Mount Desert Rock from the ship is due North, distant 10 miles, which being reversed 
gives the 1st Course diie South, on which the ship makes no Departure; and therefore the whole Distance 
will be the Difference of Latitude = 10 miles South. Answer. 



By the Table of Points opposite to the Compass, on page 10, 

2d Course W.S.W. ... 6 points = 67° 30' gives 22.96101 the number of miles in a degree of longitude 

in latitude equal to it. 

Complement of the Course 2 " =22 30 " 55.43268 the number of miles in a degree of longitude 

in latitude equal to it. 

BY CASE I. PLANE SAILING. 

1st. To find the Difference of Latitude on the 2d Course. 

25 X 22.96101 

— = 9.56712 the number of mUes in the Difference of Latitude = 9.6 Answer. 

bO 

2d» To find the Departure on the 2d Course. 

25 X 55 43268 

-T^ = 23.09695 the number of miles in the Departure = 23.1 Answer. 



By the Table of Points opposite to the Compass, on page 10, 

5d Course S.W 4 points = 45° gives 42.42637 the number of miles in a degree of longitude in 

latitude equal to it. 
Complement of the Course 4 " =45 " 42.42637 the number of miles in a degree of longitude in 

latitude equal to it. 

BY CASE I. PLANE SAILING. 

\st. To find the Difference of Latitude on the Zd Course. 

30 X 42.42637 

■ = 21.21319 the number of miles in the Difference of Latitude = 21.2 Answer. 



2d. To find the Departure on the ?>d Course. 

X 42.42637 
— ■ = 21.21319 the number of miles in the Departure = 21.2 Answer. 



TRAVERSE SAILING. 



39 



On the 4th Course, which is due West, the ship makes no Difference of Latitude, and therefore the whole 
Distance sailed will be the Departure = 20 miles. Answer. 



Having calculated all the Differences of Latitude and Departures on the several 
Courses and Distances sailed, arrange them, according to the Rule, in the form 
of a 

TRAVERSE TABLE. 



Course. 



Distance. 



Difference of Latitude. 

N. S. 



Departure. 
E. W. 



South 

w.s.w 
s.w. 

West 



10 
25 
30 

20 



10.0 

9.6 

21.2 



40.8 
Difference of Latitude made good 



23.1 
21.2 
20.0 



64.3 
Departure made good 



BY case VI. plane SAILING. 

l5^. To find the Distance made good. 



l/(40.8)2 + (64.3)2 = -j/l664.&4 + 4134. 



-1/^5799.13 = 76.15202 the number of miles in the Distance made 
good = 76.2 Answer. 



2d. To find the Course made good. 

40.8 X 60 

" ^ - = 32.14623 the number of miles in a degree of longitude in latitude equal to the Course made 

* '^•■^^^ good, viz. S. 57° 36' W., because the Difference of Latitude made good, is to the South- 

ward, and the Departure made good, is to the Westward, as per Example IX., page 21 . 
Answer. 

Tlierefore, Mount Desert Rock bears from the ship, on the point of the compass di- 
rectly opposite, viz. N. 57° 36' E. 

Part III. 

To find the direct Distance and Course from where the ship now is, to Cape Cod, the 

place bound to. 

Latitude of Mt. Desert Rock 43° 50' N. 
Cape Cod light . 42 03 N. 



Difference of Latitude . . 1 47 
Difference of Latitude made good . 



107 

40.! 



miles S. 



Departure 84 miles W. 

Departure made good . 64.3 " W. 



Dtfiference of Latitude yet to make . . 66.2 



S. 



Departure yet to make 19.7 



W. 



|/(66.2)2 + (19.7)2 = |/4382.44 



BY CASE VI. PLANE SAILING. 

l5^. To find the direct Distance yet to make. 

388.09 = y^i 



70.53 



06902 the number of miles in the direct Dis- 
tance from the ship to Cape Cod light == 
69.1 miles. Answer. 



66.2 X ' 



2d. To find the direct Course yet to make. 



57.50771 the number of miles in a degree of longitude in latitude equal to the direct Course, 
viz. S. 16° 34' W., because the Difference of Latitude yet to make, is to the Southward, 
and the Departure yet to make, is to the Westward, as per Example X., page 21. Answer 



EXAMPLE in. 

A ship in the latitude of 37° 10'' N. is bound to a port in the latitude of 
33° 00^ N., which lies 180 miles West of the meridian of the ship ; but by reason 
of contrary winds she sails on the following courses, viz. 1st, S.W. by W. 27 miles; 
2d, W.S.W. ^ W. 30 miles; 3d, W. by S. 25 miles ; 4th, W. by N. 18 miles ; 5th, 
S.S.E. 32 miles; 6th, S.S.E.f B. 27 miles; 7th, S. by E..25 miles; 8th, South 31 
miles; and 9th, S.S.E. 39 miles: required, 1st, the direct Distance and Course from 
the place left, to the port bound to ; 2d, the Distance and Course made good by the 
.ship, from the place sailed from, to where she has arrived ; and 3d, the direct Dis- 
tance and Course from where she now is, to the port bound to. 



40 TRAVERSE SAILING. 

Latitude sailed from . . 37° WN. 
" of port bound to 33 00 N. 

Difference of Latitude . . 4 10 «= 250 miles. Departure 180 miles. 

Part I * 
Tojind the direct Distance and Course from the place lefty to the port hound to. 

BY CASE VI. PLANE SAILING. 

\st. To find the Distance. 



•j/(250)2 + (180)2 = |/62500 + 32400 = |/94900 = 308.05844 the number of miles in the direct Distance 

from the place left, to the port bound to. 

An?. 

2d. To find the Course. 

250 X 60 

' tine AtiQ ' ^ 48.69213 the number of miles in a degree of longitude in latitude equal to the direct Course, 
<}U».U{)» yi2. s. 350 45/ Yf ^ because the Difference of Latitude, is to the Southward, and the De- 

parture to the Westward, as per Example XI., page 21. Answer. 

Part II. 

To find the Distance and Course made good hy the ship, from the place sailed from, to 

where she has arrived. 
By the Table of Points opposite to the Compass, on page 10, 

1st Course S.W. by W. . . 5 points = 56° 15' gives 33.33418 the number of miles in u degree of longitude in 

latitude equal to it. 
Complement of the Course 3 " =33 45 " 49.88810 the number of miles in a degree of longitude in 

latitude equal to it. 

BY CASE I. PLANE SAILING. 

\st. To find the Difference of Latitude on the \st Course. 

27 y 33 33418 

• —^ = 15.00038 the number of miles in the Difference of Latitude «= 15. Answer. 

oO 

2d. To find the Departure on the \st Course. 

27 V 49 88810 

— ^ = 22.44965 the number of miles in the Departure = 22.4 Answer, 

By the Table of Points opposite to the Compass, on p^age 10, 

2d Course W.S.W. i W. . . 6^ points = 73° 07-i-' gives 17.41706 the number of miles in a degree of longitude 

in latitude equal to it. 

Complement of the Course 1^ " = 16 52^ " 57.41635 the number of miles in a degree of longitude 

in latitude equal to it. 

BY CASE I. PLANE SAILING. 

1st. To find the Difference of Latitude on the 2d Course. 

30 X 17.417C 



60 



= 8.70853 the number of miles in the Difference of Latitude = 8.7 Answer. 



2d. To find the Departure on the 2d Course. 

— ^ • = 28.70818 the number of miles in the Departure = 28.7 Answer. 

60 

By the Table of Points opposite to the Compass, on page 10, 

3d Course W. by S. . . . 7 points = 78° 45' gives 11.70539 the number of miles in a degree of longitude in 

latitude equal to it. 
Complement of the Course 1 point = 11 15 " 58.84702 the number of miles in a degree of longitude in 

latitude equal to it. 

BY CASE I. PLANE SAILING. 

l5^. To find the Difference of Latitude on the 3tZ Course. 

25 X 11 70539 

— '. = 4.87725 the number of miles in the Difference of Latitude = 4.9 Answer. 

60 

2d. To find the Departure on the 2>d Course. 

25 V 58 84702 

— — '■ = 24.51959 the number of miles in the Departure = 24.5 Answer. 



* Bowditch has omitted to give this part of the Example. Why? 



TRAVERSE SAILING. 41 

By the Table of Points opposite to the Compass, on page 10, 

4tli Course W. by N. . . . 7 points = 78° 45' gives 11.70539 the mnnber of miles in a degree of longitude in 

latitude equal to it. 
Complement of the Course 1 point = 11 15 " 68.84702 tlio number of miles in a degree of longitude in 

latitude equal to it. 

BY CASE I. PLANE SAILING. 

1.?^. Tojiiid the Difference of Latitude on the 4ih Course. 

18 X 11 70539 

. '- = 3.51162 the number of miles in the Difference of Latitude = 3.5 Answer. 

60 

2d. Tojind the Departure on the Ath Course. 

18 y '^8 8470*2 

. ^^-^ = 17.65411 the number of miles in the Departure •■= 17.7 Answer. 



By the Table of Points opposite to the Compass, on page 10, 

6th Course S.S.E 2 points ^ 22° 30' gives 55.43268 the number of miles in a degree of longitude 

in latitude equal to it. 
Complement of the Coiirse 6 " =67 SO " 22.96101 the number of miles in a degree of longitude 

in latitude equal to it. 

BY CASE I. PLANE SAILING. 

\st. Tojind the Difference of Latitude on the 5t7i Course. 

32 X 55.43268 



60 



= 29.56410 the number of miles in the Difference of Latitude = 29.6 Answer. 



2d. To find the Departure on the 5th Course. 

32 X 22.96101 

• — = 12.24587 the number of miles in the Departure = 12.2 Answer. 



By the Table of Points opposite to the Compass, on page 10, 
6th Course S.S.E. :J E. . . 2f points = 30° 56^' gives 51.45361 the number of miles in a degree of longitude 

in latitude equal to it. 
Complement of the Course 5^ " = 59 03J " 30.84612 the number of miles in a degree of longitude 

in latitude equal to it. 

BY CASE I. PLANE SAILING. 

1st. To find the Difference of Latitude on the 6th Course. 

27 X 51.45361 

= 23.15412 the number of miles in the Difference of Latitude = 23.2 Answer. 

2d. To find the Departure on the Qth Course. 

27 X 30.84612 

• — = 13.88075 the number of miles in the Departure = 13.9 Answer. 



By the Table of Points opposite to the Compass, on page 10, 

7th Course S. by E. . . . 1 point ^ 11° 15' gives 58.84702 the number of miles in a degree of longitude in 

latitude equal to it. 
Complement of the Course 7 points = 78 45 " 11.70539 the number of miles in a degree of longitude in 

latitude equal to it. 

BY CASE I. PLANE SAILING. 

Jst. To find the Difference of Latitude on the 7th Course. 

25 X 58.84702 

= 24.51959 the number of miles in the Difference of Latitude = 24.5 Answer. 

60 

2d. To find the Departure on the 1th Course. 

25 X n 70539 

• — -^ = 4.87725 the number of miles in the Departure = 4.9 Answer. 



On the 8th Course, which is due South, the ship makes no Departure ; and therefore the whole Distance 
sailed, viz. 31 miles, will be the number of miles in the Difference of Latitude = 31. Answer. 



By the Table of Points opposite to the Compass, on page 10, 

9th Course S.S.E 2 points = 22° 30' gives 55.43268 the number of miles in a degree of longitude 

in latitude equal to it. 

Complement of the Course 6 " «= 67 30 " 22.96101 the number of miles in a degree of longitude 

in latitude equal to it. 

BY CASE I. PLANE SAILING. 

Is^. To find the Difference of Latitude on the 9th Course. 

39 X 55.43268 

= 36.03124 the number of miles in the Difference of Latitude = 36. Answer. 



2cL To find the Departure on the 9th Course. 

X 22.96101 

= 14.92466 the number of miles in the Departure = 14.9 Answer. 



60 



42 



TRAVERSE SAILING. 



Having calculated all the Differences of Latitude and Departures, on the several 
Courses aud Distances sailed, arrange them, according to the Kule, in the form 
of a 

TRAVERSE TABLE. 









1 






Difference of Latitude. 


Departure. 


Course. 


Distance. 


N. 


S. 


E. 


W. 


S.W. by W. 


27 




15.0 




22.4 


W.S.AV. i VV. 


30 




8.7 




28.7 


W. by S. 


25 




4.9 




24.5 


W. by N. 


18 


3.5 






17.7 


S.S.E. 


32 




29.6 


12.2 




S.S.E. f E. 


27 




23.2 


13.9 




S. by E. 


25 




24.5 


4.9 




South 


31 




31.0 






S.S.E. 


39 




36.0 


14.9 








3.5 


172.9 
3.5 


45.9 


93.3 
45.9 


Kemainder 169.4 


Remainder 47.4 




Difference of Latitude made good. 


Departure made good. 
1 



BY CASE VI. PLANE SAILING. 

l5^. To find the Distance made good. 



|/(169.4)2 + (47.4)2 = |/28696.36 + 2246.7G = -1/30943.12 = 175.90657 the number of miles in the Distance 

made good = 175.9 Answer. 



2d. To find the Course made good. 



175.90657 



57.78011 the number of miles in a degree of longitude in latitude equal to the Course made 
good, viz. S. 15° 38' W., because the Difference of Latitude made good, is to the Sonth- 
Avard, and the Departure made good, is to the WestAvard, as per Example XII., page 21. 
Answer. 

Therefore, the place sailed from, bears from the ship, on the point of the compass di- 
rectly opposite, viz. N. 15° 38' E. 



Part III. 

To find the direct Distance and Course from where the ship now is, to the port 

bound to. 

Latitude sailed from . . . 37° 10' N. 
" of port bound to . 33 00 N. 



Difference of Latitude . . 4 10 ^ 
Difference of Latitude made good . 

Difference of Latitude yet to make . 



250 miles S. 
169.4 " S. 



Departure 180 miles W. 

Departure made good . 47.4 " W. 



Departure yet to make 132.1 



W. 



BY CASE YI. PLANE SAILING. 

1st. To find the direct Distance yet to make. 



1// (80.6)2 -j- (132.6)2 = i/ 6496.36 + 17582.76 = i/ 24079.12 = 155.17449 the number of miles in the direct 

Distance from the ship, to the port 
bound to = 155.2 Answer. 



80.6 X 60 
155.17449 '' 



2d. To find the direct Course yet to make. 



31.16492 the number of miles in a degree of longitude in latitude equal to the direct Course, 
viz. S. 58° 42' W., because the Difference of Latitude yet to make, is to the Southward, 
and the Departure yet to make, is to the Westward, as per Example XIII., page 22. 
Answer. 

»— 



The Latitude sailed from 37° 10' N. 

169 
Difference of Latitude made good is 169.4 miles = -— = 2 49 S. 

oO 



The Latitude the ship is in 34 21 N. Answer. 



PARALLEL SAILING. 



Tn the problems of Plane Sailing and Traverse Sailing, the Earth has been con- 
sidered as constituting an extended plane ; but in the solution of problems in which 
longitude is concerned, we must necessarily take into consideration the real spheri- 
cal form of the Earth, with the meridians meeting each other at the poles. The 
Departure made on any parallel of latitude will not give the same Difference of Lon- 
gitude that it would if made in a higher or lower latitude. Hence, the Difference 
of Longitude cannot be found by either of the two former sailings. The distance 
between any two meridians measured on a parallel of latitude is called the meri- 
dian distance, and decreases in proceeding from the Equator, where it is greatest, 
towards the poles, where it is nothing. 

When a ship sails due East or due West on the surface of the Earth, she keeps in 
the same latitude and describes a circle parallel to the Equator, and this is called 
parallel sailing. 

In Parallel Sailing, the distance sailed East or West, is the Departure made good 
by the ship, and is equal to the distance between the meridian sailed from and 
the meridian arrived at, in the latitude of the parallel sailed on. 

The Longitude of a place is its distance east or west from any given principal or 
first meridian, fixed at pleasure, such as that of Washington, Greenwich, Paris, &c., 
and is measured by that part of the Equator intercepted between the principal me- 
ridian and the meridian that passes through the given place, which is equal to the 
angle at the pole, between the two meridians. 

As Longitude is counted from the principal or first meridian. East and West, on 
the Equator, until it comes to the opposite meridian, therefore a ship's longitude 
cannot exceed 180° East or West. 

The Difference of Longitude between two places, if both East or both West, is 
found by subtracting the less longitude from the greater ; but if one longitude be 
East and the other West, the Difference of Longitude is found by adding both lon- 
gitudes together. Should their sum exceed 180°, subtract it from 360°, and the 
remainder will be the Difference of Longitude. 

In East longitude sailing to the Eastward, or in West longitude sailing to the 
Westward, the Difference of Longitude must be added to the Longitude left, to get 
the Longitude in, because the ship sails farther away from the principal or first 
meridian. 

But in East longitude sailing to the Westward, or in West longitude sailing to 
the Eastward, the Difference of Longitude must be siibtracted from the Longitude 
left, to get the Longitude in, because the ship sails nearer to the principal or first 
meridian. 

When the Longitude decreases and the Difference of Longitude is greater than 
the Longitude left, the Longitude left, must be subtracted from the Difference of Lon- 
gitude, to get the Longitude in, which will be of a different name from the Longitude 
left, because the ship in this case, crosses the principal or first meridian. 

This sailing is particularly useful in making small or low islands, in which case 
it is usual to run into the latitude and then steer due East or due West. 

(43) 



44 PARALLEL SAILING. 

CASE I. 

Given the Difference of Longitude between two places in the same parallel of 
Latitude, to find the Distance between them. 

RULE. 
Multiply the Difference of Longitude in miles, by the number of miles in a degree 
of longitude in the latitude of the parallel on which the ship sailed, and divide the 
product by 60 ; the quotient will be the number of miles in the Distance. 

EXAMPLE. 
Suppose a ship, in the latitude of 49° 30^ North or South, sails directly East or 
West, until her Difference of Longitude be 3° 30^ (— 210 miles) ; required the Dis- 
tance sailed. 

By the Table of Miles in a degree of Longitude in every Latitude, on page 9, 

Latitude 49° 30' gives 38.96685 the number of miles in a degree of longitude. 

210 X 38.96685 

— = 136.38398 the number of miles in the Distance sailed = 136.4 Answer. 

w 

CASE II. 

Given the Distance between two places on the same parallel of Latitude, to find 
their Difference of Longitude. 

RULE. 

Multiply the Distance in miles by 60, and divide the product by the number of 
miles in a degree of longitude in the latitude of the parallel on which the ship sailed; 
the quotient will be the number of miles in the Difference of Longitude, which divided 
by 60 gives the number of degrees and minutes to be added to, or subtracted from, 
the Longitude left, accordingly as we are increasing or lessening our longitude. 

EXAMPLE. 
Suppose a ship, in the latitude of 49° 30^ North or South and in longitude 36° 40^ 
West, sails directly West, the Distance 136.4 miles ; required the Difference of Lon- 
gitude and the Longitude in. 

By the Table of Miles in a degree of Longitude in every Latitude, on page 9, 

Latitude 49° 30' gives 38.96685 the number of miles in a degree of longitude. 

136.4 X 60 210 

■ o^^^V^. = 210.02467 the number of miles in the Difference of Longitude = -— = 3^= 30' W. 
38.96685 ^ 60 

Which added to the Longitude left 36 40 W. 



Gives the Longitude the ship is in 40 10 W. Answer. 

CASE III.* 

Given the Distance between two places on the same parallel and their Difference 
of Longitude, to find their Latitude. 

RULE. 

Multiply the Distance in miles by 60, and divide the product by the Difference of 
Longitude in miles ; the quotient will be the number of miles in a degree of longi- 
tude in the latitude of the parallel on which the places are. 

EXAMPLE. 
Suppose a ship having sailed due East or West the Distance 136.4 miles, found 
she had made a Difference of Longitude of 3° 30^ (= 210 miles) ; on what parallel 
of Latitude did she sail ? 

. — = 38.97143 the number of miles in a degree of longitude in the latitude of the parallel On which 

210 tlie ship sailed, viz. 49° 30' N. or S., as per Example XIV., page 22. Answer. 

REMARK. 

The several Cases of Parallel Sailing may be solved by other rules, as follows : — 

Case I. Rule. — Multiply the number of miles in a degree of longitude in the 
latitude the ship sailed, by the number of degrees and minutes (turned into the deci- 
mal of a degree) in the Difi'erence of Longitude, and the product will be the miles 
in the Distance sailed. 

Example the same as before. 38.96685 X 3.5 = 136.38398, the number of miles in the Distance sailed = 136.4 

Answer. 

* Bowditch has omitted to give this Case! Why? Question IT. for Exercise cannot be solved without it. 



PARALLEL SAILING. 45 

Case II. Rule.— Divide the Distance in miles, by the number of miles in a de- 
gree of longitude in the latitude of the parallel on which the ship sailed, and the 
quotient will be the Diifference of Longitude in degrees and decimal of a degree. 

Example the same as befo-e. - = 3°.50041 = 3° 30' W. the Difference of Longitude. 

^ 38.96685 

"Which added to 36 40 W. the Longitude left. 



Gives .... 40 10 W. the Longitude the ship is in. Answer. 

Case III. Rule. — Divide the Distance in miles, by the number of degrees and 
minutes (turned into the decimal of a degree) in the Difference of Longitude be- 
tween the two places, and the quotient will be the miles in a degree of longitude, in 
the latitude of the parallel on which the places are. 

Example the same as before. — '- = 38.97143 the number of miles in a degree of longitude in the latitude 
S.5 of the parallel on which the ship sailed, viz. 49° 30' N. or S., 

as per Example XIV., page 22. Answer. 

QUESTIONS 

To exercise the Learner in the foregoing Rides. 

QUESTION L 

A ship in the latitude of 32° North or South, sails due East or "West, till her Differ- 
ence of Longitude is 384 miles ; required the Distance sailed. 

BY CASE I. 

By the Table of Miles in a degree of Longitude in every Latitude, on page 9, 

Latitude 32° gives 50.88282 the number of miles in a degrea of longitude. 

384 V 50 88282 

'- = 325.65005 the number of miles in the Distance sailed = 325.7 Answer. 

60 

QUESTION IL 

A ship from the latitude of 53° 36^ North or South and longitude 10° 18^ East sails 
due West, the Distance 236 miles ; required the Difference of Longitude and the 
Longitude in. 

BY case II. 

Latitude 53° 36' gives 35.60505 the number of miles in a degree of longitude, as per Example Y., page 14. 

OQfi v/ CQ 3Q8 

- = 397.69639 the number of miles in the Difference of Longitude = -^ = 6° 38' W. 

35.60505 60 

Which subtracted from the Longitude left 10 18 E. 

Gives the Longitude the ship is in ... 3 40 E. Answer. 

QUESTION IIL 

If two ships in the latitude of 44° 30^ North or South, Distant 216 miles, should 
sail directly South or North, until they arrive in the latitude of 32° 17^ North or 
South, what Distance would they be from each other ? 

BY case II. 
By the Table of Miles in a degree of Longitude in every Latitude, on page 9, 

Latitude 44° SO' gives 42.79500 the number of miles in a degree of longitude. 

216 X 60 

= 302.83912 the number of miles in the Difference of Longitude. 

42.79500 

BY CASE I. 

By the Table of Miles in a degree of Longitude in every Latitude, on page 9, 

Latitude 32° 17' gives 50.72495 the number of miles in a degree of longitude, as per Example VT., page 14. 

Vl9 R'^Q12 V 50 72495 

"^ ^ = 256.02499 the number of miles that the ships are Distant from each other = 256. 

60 Answer. 

QUESTION IV. 

A ship having run due East or West for three days, at the rate of 5 knots an hour 
(sailed a Distance of 360 miles) finds she has made a Difference of Longitude of 
8° 16'' (= 496 miles) ; on what parallel of Latitude did she sail ? 

BY CASE III. 

— — = 43.54839 the number of miles in a degree of longitude in the latitude of the parallel on which 

*96 tlie ahip sailed, viz. 43° 28' North or South, as per Example XV., page 22. Answer. 



MIDDLE LATITUDE SAILI^s'G. 



"When a ship sails due North or due South, she keeps on the same meridian, and 
therefore does not change her longitude, and her distance run is the difference of 
latitude : consequently her place is easily determined by the latitude left and the 
difference of latitude. Again, vrhen a ship sails due East or due West, her differ- 
ence of longitude is found by the latitude in, and departure or meridian distance, 
as already explained in Parallel Sailing ; but T^vhen she sails upon any other course, 
she changes both her latitude and longitude. Now the difference of longitude can- 
not be found either from the departure, considered as a meridian distance in the 
latitude left or that come to ; for in the greater latitude it would give the difference 
of longitude too much, and in the less latitude too little : the departure is therefore 
taken as a meridian distance in the mean or middle of the two latitudes, and then 
the difference of longitude can be found as in Parallel Sailing. Hence this method, 
it is evident, is compounded of Plane Sailing and Parallel Sailing. 

Middle Latitude Sailing, therefore, is that in which the Earth and Sea are consi- 
dered as constituting a sphere or globe ; the meridians as great oircles on its surface 
meeting each other at the poles ; the ship as sailing at any time along a curved 
line (called the " Loxodromic" or oblique curve) cutting the meridians in the same 
constant angle, while on the same tack ; and the departure in going from one place 
to another as equal to the distance between their meridians on the middle par- 
allel of latitude between the two places. 

The Middle Latitude is half the sum of the two latitudes when they are of the 
same name, that is, both North or both South ; but half their difference when of 
contrary names, that is, one North and the other South. 

This method of sailing, although not strictly accurate, especially in high latitudes, 
approaches sufficiently near the truth for a day's run or the distance a sliip may sail 
in twenty-four hours. It is used principally in low latitudes, or when the two latitudes 
do not differ much from each other, and the difference of longitude is small, because 
the Meridional Parts, in Table III. of the Epitome, are calculated to whole miles 
only, which renders them inapplicable to day's work.* 

* Bo-^ditch says, " Middle Latitude Sailing is only an approximation, but it is very mnch used in calculat- 
ing short runs and days' works; but in crJculating large distances across distant parallels, it is liable to 
error." Yet notwithstanding this, he gives examples in several instances, as if he did not know they were 
unsuitable or improper, and in opposition to the well-known fact above stated. See Example of Case I., 
Questions for Exercise T., TI., Y., YIII., and IX. These properly require to be solved by Mercator's Sailing, 
which is perfectly accurate. 

(46) 



MIDDLE LATITUDE SAILING. 47 

CASE I. 

Given the Latitudes and Longitudes of two places, to find their Distance and Bear- 
ing (or Course). 

RULE. 

Ist. Tofnd the Departure. 

Multiply the Difi'erence of Longitude in miles, by the number of miles in a degree 
of longitude in the Middle Latitude, and divide the product by 60 ; the quotient 
will be the number of miles in the Departure. 

2d. Tojind the Distance. 

Add the square of the Difi'erence of Latitude in mJles, to the square of the Depart- 
ure in miles, and from the sum, extract the square root ; which will be the number 
of miles in the Distance. 

Zd. Tojind the Course. 

Multiply the Difi'erence of Latitude in miles, by GO, and divide the product by the 
Distance in miles ; the quotient will be the number of miles in a degree of longi- 
tude in latitude equal to the Course. 



EXAMPLE. 

Required the Distance and Course between Cape Cod lighthouse, in the Latitude 
of 42° 03^ N., Longitude 70° 04^ W., and the Island of St. Marv (one of the Azores 
or Western Islands), in the Latitude of 36° 59^ N., and Longitude 25° 10^ W. 

Be7nark.—The latitude of Cape Cod (Hio-hknd) light is 42° 02^ N., and the lon- 
gitude is 70° 04^ W. The latitude of the Island of St. Mary is 36° 50' N., and the 
longitude is 25° 16^ W. But as we want to work out all the"^ Examples in Bowditch 
by our method, we must use his figures, although knowing them to be not strictly 
correct. 

Cape Cod's Latitude... 42° 03' N 42° 0.3' Cape Cod's Longitude... 70° 04' W. 

St. Mary's " ...36 59 N 36 59 St. Blary's "" ... 25 10 W. 



Difference of Latitude 5 04 2)79 02 Difference of Longitude 44 54 

60 60 

39 31 

Dif. of Lat. in miles... 304 Middle Latitude. Dif. of Long, in miles... 2(i94 

Is^ Tojind the Departure. 

By the Table of Miles in a degree of Longitude in every Latitude, as per Example \l'i., page 14, 

The Middle Latitude 39° 31' gives 46.28633 the number of miles in a degree of longitude. 

2694 X 46.28633 

~ = 2078.25622 the number of miles in the Departure = 2078. Answer. 

2d. Tojind the Distance. 



F (304;2 + (2078.256)!" == |/92416 + 4319148.001536 = -,/4421564.001536 = 2102.75153 the number of miles 

in the Distance = 
2102.8 Answer. 

3c?. Tojind the Course. 

304 X 60 

2102.752 = ^'^"'^^^ t^®cioT/\^'* ""^ ™"^' '"^ \ '^^^4^ «f longitude, in latitude equal to the Course, viz. 
S. 81° 41' L., as per Example X\I., page 22. Answer. 

Because the Island of St. :\Iary is in a less North latitude than Cape Cnd. therefore 
St. Mary is to the Southward of Cape Cod ; and because the Island of St. Mary is in a less 
West longitude than Cape Cod, therefore St. Mary is to the Eastward of Care Cod 
Hence, Cape Cod lighthouse bears from the Island of St, Mary, on the point of the 
compass directly opposite, viz. N. 81° 41' W. » j> i-^ ^<' "^ 



48 MIDDLE LATITUDE SAILING. 

CASE II. 

Given both Latitudes, one Lonj^itude, and the Departure from the meridian, to 
find the Distance, Course, and Difference of Longitude. 

RULE. 

Isi. Tojind the Distance. 

Add the square of the Difference of Latitude in miles, to the square of the Depart- 
ure in miles, and from the sum extract the square root, which will be the Distance 
in miles. 

2d. To find the Course. 

Multiply the Difference of Latitude in miles, by 60, and divide the product by the 
Distance in miles ; the quotient will be the number of miles in a degree of longitude 
in latitude equal to the Course. 

Zd. To find the Difference of Longitude. 

Multiply the Departure in miles, by 60, and divide the product by the number of 
miles in a degree of longitude in the Middle Latitude ; the quotient will be the Dif- 
ference of Longitude in miles. 



EXAMPLE. 

A ship in the Latitude of 49° 57^ North, and Longitude 15° 16^ West, sails South- 
Westerly till her Departure is 194 miles and Latitude in, is 47° 18^ North. Required 
the Distance, Course, and Longitude in. 

Latitude left 49° 57' N 49° 57' 

in 47 18 N 47 18 



Difference of Latitude 2 39 2)97 15 

60 

48 38 Middle Latitude. 

Dif. of Lat. in miles... 159 



\st. To find the Distance. 



|/(159)2 + (194)3 = |/25281 + 37636 ==^ |/62917 = 250.83261 the number of miles in the Distance = 250.8 

Answer. 

2d. To find the Course. 

169 X 60 

_—-__- == 38.03327 the number of miles in a degree of longitude, in latitude equal to the Course, viz. 
250-833 g_ 50O 40^ ^ff ^ ^s per Example XVII., page 23. Answer. 

Because the place bound to, is in a less North latitude than the place left and to 
the Westward of it. 



Zd. To find the Difference of Longitude. 

By the Table of Miles in a degree of Longitude in every Latitude, as per Example VIII., page 14. 
The Middle Latitude 48° 38' gives 39.65247 the number of miles in a degree of Longitude. 

194 X 60 294 

■ = 293.54540 the number of miles in the Difference of Longitude = — = 4° 54' W. 

Which added to the Longitude left 15 16 W. 



Gives the Longitude in 20 10 W, 

Answer. 



MIDDLE LATITUDE SAILING. 49 

CASE III. 

Given one Latitude, the Course, and Distance, to find the Difference of Latitude 
and Difference of Longitude. 

RULE. 

Isi. To find the Difference of Latitude. 

Multiply the Distance in miles, by the number of miles in a degree of longitude, 
in latitude equal to the Course, and divide the product by 60 ; the quotient will be 
the Difference of Latitude in miles. 

2cZ. To find the Departure.* 

Multiply the Distance in miles, by the number of miles in a degree of longitude, 
in latitude equal to the Complement of the Course, and divide the product by 60 ; 
the quotient will be the Departure in miles. 

Zd. To find the Difference of Longitude. 

Multiply the Departure in miles, by 60, and divide the product by the number of 
miles in a degree of longitude, in the Middle Latitude ; the quotient will be the Dif- 
ference of Longitude in miles. 



EXAMPLE. 

A ship in the Latitude of 42° 30^ North, and Longitude 58° bV West, sails 
S.E. by S. 300 miles. Eequired the Latitude and Longitude in. 

\st. To find the Difference of Latitude. 

By the Table of Points opposite to the Compass, on page 10, 

The Course S.E. by S. 3 points = 33° 45' gives 49.88810 the number of miles in a degree of longitude in lati- 
tude equal to it. 

300 X 49.88810 249 

— = 249.44050 the number of miles in the Difference of Latitude = — = 4^ 09' S. Answer. 

W) 60 

Latitude left 42° 30' N 42° 30^ 

Diflerence of Latitude 4 09 S. 



Latitude in 38 21 N 38 21 

Answer. 

2)80 61 

40 26 Middle Latitude. 

2d. To find the Departure. 

By the Table of Points opposite to the Compass, on page 10, 

The Complement of the Course, yiz. 5 points = 56° 15' gives 33.33418 the number of miTes fn a degree of Ion.- 

gitude in latitude equal to it. 

300 X 33.33418 

• — = 166.67090 the number of mil^ in the Departure = 16S.T Answer. 

3(f. To find the Difference of Longitude. 

By the Table of Miles in a degree of Longitude in every Latitude, as per Example IX., page 15, 
The Middle Latitude 40° 26' gives 45.66964 the number of miles in a degree of longitude. 

= 218.96954 the number of miles in the Difference of Longitude = — - = 3"= 39^ EL 



Whicli subtracted from the Longitude left 58 51 W. 



Gives the Longitude in. 55 12 W» 

Answer. 



* The Departure may be found otherwise, by Case IT. of Plane Sailing, thus ; 



|/(Di8tan<-e)2 — (Dif. of Lat.)2 == -|/(300y^ — (249.441)2 = |/'90000 — 62220.812481 = |^/ 27779.187519 =- 

166.67090 the number of miles in the Departure = 166,7 Answer. 

4 



50 MIDDLE LATITUDE SAILING. 

CASE IV. 

Given both Latitudes and the Course, to find the Distance, Departure, and Differ- 
ence of Longitude. 

RULE. 

1st. To find the Distance. 

Multiply the Difference of Latitude in miles, by 60, and divide the product by the 
number of miles, in a degree of longitude in latitude equal to the Course ; the quo- 
tient will be the Distance in miles. 

2d. To find the Departure.'^ 

Multiply the Distance in miles, by the number of miles in a degree of longitude, 
in latitude equal to the Complement of the Course, and divide the product by 60 ; 
the quotient vs^ill be the Departure in miles. 

Zd. To find the Difference of Longitude. 

Multiply the Departure in miles, by 60, and divide the product by the number of 
miles in a degree of longitude, in the Middle Latitude ; the quotient will be the Dif- 
ference of Longitude in miles. 

EXAMPLE. 

Suppose a ship sailing from a place in the Latitude of 40° 57^ North, and Longi- 
tude 30° West, makes a Course of S. 39° TV., and then, by observation, is in the Lat- 
itude o«f 47° 44^ North ; required the Distance run, Departure, Difference of Longi- 
tude, and the Longitude in. 

Latitude left 49° 57' N 49° 57' 

in 47 44 N 47 44 



Difference of Latitude 2 13 2)97 41 

60 

48 61 Middle Latitude. 

Dif. of Lat. in miles... 133 

\st. To find the Distance. 

By the TtCble of Miles in a degree of Longitude in every Latitude, on page 9, 

The Course S. 39° W. gives 46.62882 the number of miles in a degree of longitude. 

^ = 171.13871 the number of miles in the Distance= 171.1 Answer. 

46.62882 

2d. To find the Departure. 

By the Table <rf" Miles in a degree of Longitude in every Latitude, on page 9j 

The Complement of the Course 51° gives 37.75922 the number of miles in a degree of longitude. 

171 139 "^ 37 75922 

. — — '■ = 107.70125 the number of miles in the Departure = 107.7 Answer. 

60 

Zd. To find the Difference of Lmi;/>fu'Jf. 

By the Table of Miles in a degree of Longitude in every Latitude, as per Example X., page 15, 

The Middle Latitude 48° 51' gives 39.48194 the number of miles in a degree of longitude. 

107 701 V 60 164 

^ = 163.67129 the number of miles in the Difference of Lonsritude = — - -= 2° 44' W. 

39.48194 60 

Which added to the Longitude left .30 00 W, 



Gives the LongitTide in 32 44 W. 

AnsAver. 



* The Departure may be found otherwise, by Case IV. of Plane Sailivg, thus :— 



T/(Distance)2 — (Dif. of Lat.)2 = i/''(171 .13871)2— (13S)2 = -j/292'!8,45^n604641 — 17689 == |/ 11599.4580604641 

= 107.70078 the number of miles in the Departure ^ 107.7 Answer. 



MIDDLE LATITUDE SAILING. 51 

CASE V. 

Given both Latitudes and the Distance, to find the Course, Departure, and Differ- 
ence of Longitude. 

RULE. 

1st. Tojind the Course. 

Multiply the Difference of Latitude in miles, by 60, and divide the product by the 
Distance in miles ; the quotient will be the number of miles in a degree of longitude, 
in latitude equal to the Course. 

2d. To find tlie Departure.* 

Multiply the Distance in miles, by the number of miles in a degree of longitude, 
in latitude equal to the Complement of the Course, and divide the product by 60 ; 
the quotient vrill be the Departure in miles. 

Zd. To find the Difference of Longitude. 

Multiply the Departure in miles, by 60, and divide the product by the number of 
miles in a degree of longitude, in the Middle Latitude ; the quotient will be the Dif- 
ference of Longitude in miles. 



EXAMPLE. 

Suppose a ship sails 300 miles North-Westerly from a place in the Latitude of 
37° North, and Longitude 32° 16^ West, until she is in the Latitude of 41° North ; 
required her Course, Departure, Difference of Longitude,, and Longitude in. 

Latitude in 41° 00' N... 41° 00' 

left 37 00 N 37 00 



240 X 



Difference of Latitude 4 00 2)78 00 

60 

39 00 Middle Latitude. 

Dif. of Lat. in miles... 240 

l5^. To find tlie Course. 

48.00000 the number of miles in a degree of longitude, in latitude equal to tlie Course, vie. 
N. 36° 52' W., as per Example XVIII., page 23, Answer. 

2d. To find the Departure. 

By the Table of Miles in a degree of Longitude in every Latitude, as per Example XI., page 15, 

The Complement of the Course 53° 08' gives 35.99741 the number of miles in a degree of longitude. 

300 X 35 99741 

• —-^ — - = 179.98705 the number of miles in the Departure = 180. Answer. 

60 

3c?. To find the Difference of Longitude. 

By the Table of Miles in a degree of Longitude in every Latitude, on page 9, 

The Middle Latitude .39° 00' gives 46.62882 the number of miles in a degree of longitude. 

179 987 X 60 232 

— : — —-— = 231.59968 the number of miles in the Difference of Longitude = — - = 3° 52' W. 
46.62882 60 

Which added to the Longitude left 32 16 W, 



Gives the Longitude in 36 08 W. 

Answer. 



The Departure may be found otherwise, by Case IV. of Plane Sailing, thus ; 



|/(Distance)^ — (Dif. of Lat.): = •^/(300)2 — (240)2 = -j/gOOGO — 57600 = i/32400 = 180 the number 

of miles in the Departure. Answer. 



52 MIDDLE LATITUDE SAILING. 

CASE VI. 

Given one Latitude, the Course, and Departure, to find the Distance, DijBTofence 
of Latitude, and Difference of Longitude. 

RULE. 

1st. Tojind the Distance. 

Multiply the Departure in miles, by 60, and divide the product by the number of 
miles in a degree of longitude, in latitude equal to the Complement of the Course ; 
the quotient will be the Distance in miles. 

2d. Tojind the Difference of Latitude* 

Multiply the Distance in miles, by the number of miles in a degree of longitude, 
in latitude equal to the <}ourse, and divide the product by 60 ; the quotient will be 
the Difference of Latitude in miles 

Zd. Tojind the Difference of longitude. 

Multiply the Departure in miles, by 60, and divide the product by the number of 
miles in a degree of longitude, in the Middle Latitude ; the quotient will be the Dif- 
ference of Longitude in miles. 

EXAMPLE. 

A ship in the Latitude of 50° 10^ South, and the Longitude of 30° 00^ East, sails 
E.S.E. until her Departure is 160 miles ; required her Distance sailed. Difference of 
Latitude, and Difference of Longitude. 

\st. Tojind the Distance. 

By the Table of Points opposite to the Compass, on page 10, 

Th9 Complement of the Course 2 points = 22° 30' gives 55.43268 the number of miles in a degree of longi- 
tude in latitude equal to it. 

-— -— — = 173.18304 the number of miles in the Distance = 173.2 Answer. 
55.43268 

2d. Tojind the Difference of Latitude. 

By the Table of Points opposite to the Compass, on page 10, 

The Course E.S.E. 6 points = 67° 30^ gives 22.96101 the number of miles in a degree of longitude in latitude 

equal to it. 

17*^18^ V 22Q6101 66 

1. o. o ^ . _ ^ 66.27428 the number of miles in the Difference of Latitude = 66.3 — - 1° 06' S. 



60 



Latitude left 50° 10' S 50° 10' 

Difference of Latitude 1 06 S. 



Latitude in 51 16 S 51 16 

Answer. 

2)101 26 



50 43 Middle Latitude. 

3c?. Tojind the Difference of Longitude. 

By the Table of Miles in a degree of Longitude in every Latitude, as per Example XII., page 15, 

The Middle Latitude 50° 43' gives 37.98927 the number of miles in a degree of longitude. 

== 252.70293 the number of miles in the Difference of Longitude = ^ = 4° 13' E. 
37.98927 60 

Which added to the Longitude left 30 00 E. 

Gives the Longitude in 34 13 E. Answer. 

* The Difference of Latitude may be found otherwise, by Case V. of Plane Sailing, thus : — 



^(Dl3tauce)2 — (Departure)^ = |/ (173.183)2— (160)2 = |/-29992,351489 — 25600 == |/ 4392.351489 = 66.27474 

the number of miles in the Difference of Latitude = 66.3 Answer. 



MIDDLE LATITUDE SAILING. 53 

CASE VII. 

Given one Latitude, the Distance sailed, and Departure from the meridian, to find 
the Course, Dillerence of Latitude, and DiflFerence of Longitude. 

RULE. 

Isi. Tojind th^ Course. 

Multiply the Departure in miles, by 60, and divide the product by the Distance in 
miles ; the quotient will be the number of miles in a degree of longitude, in latitude 
equal to the Complement of the Course. Subtract the Complement of the Couree 
from 90°, and the remainder will be the Course in degrees and minutes. 

2d. Tojind the Difference of Latitude.^ 

Multiply the Distance in miles, by the number of miles in a degree of longitude, 
in latitude equal to the Course, and divide the product by GO ; the quotient will be 
the Diti'erence of Latitude in miles. 

3c?. Tojind the Difference of Longitude. 

Multiply the Departure in miles, by 60, and divide the product by the number of 
miles in a degree of longitude, in the Middle Latitude ; the quotient will be the Dif- 
ference of Lon*<itude in miles. 



EXAMPLE. 

A ship in the Latitude of 49° 30^ North, and the Longitude of 25° 00^ West, sails 
South-Easterly 215 miles, until her Departure from the meridian is 167 miles ; re- 
quired the Course steered, and the Latitude and Longitude the ship is in. 

\st. Tojind the Course. 

167 X 60 

— — — = 46.60465 the number of miles in a degree of longitude in latitude equal to tfie Complement of the 
'^'■^ Course, viz. 39° 02', as per Example XIX., page 23. Answer. 

The Complement subtracted from 90*^ gives the Course S. 50° 58' E., as per Example. 

2d. To find the Difference of Latitude. 

By the Table of Miles in a degree of Longitude in every Latitude, as per Example XIII., page 15, 
The Course S. 50° 58' E. gives 37.78631 the number of miles in a degree of longitude in latitude equal to it. 
215X37.78631 ,,,,„„„, , , . ., . \^ ^.^ .^.. 135 



60 



135.40094 the number of miles in the Difference of Latitude = — =2° 15' S. 



Latitude left 49° 30' N 49° 30' 

Difference of Latitude 2 15 S. 



Latitude in 47 15 N 47 15 

Answer. 

2)96 45 

48 23 Middle Latitude. 

Zd. To find the Difference of Longitude. 

Bj the Table of Miles in a degree of Longitude in every Latitude, as per Example XlV., page 15, 



The Middle Latitude 48° 23' gives 39.84857 the number of miles in a degree of longitude. 

252 
?s in the Difference of Longitude = — 4° 12' E. 

Which subtracted from the Longitude left 25 00 W. 



— ^-f- — = 251.70288 the number of miles in the Difference of Longitude = — 4° 12' B 
39.84857 ° 60 



Gives the Longitude in 20 48 W. Answer. 



* The Difference of Latitude may be found otherwise, by Case Y. of Plane Sailing, thus : 



y^(Distance)2 — (Departure)^ = ^(215)2 — (167)2 = -j/46225 — 27889 = -|/l8o36 = 135.41049 the number 

of miles in the Difference of Latitude = 135.4 Answer 



54 MIDDLE LATITUDE SAILING. 

• CASE VIIL* 

Given one Latitude, Departure, and the Difference of Longitude, to find the other 
Latitude, Distance, and Course. 

RULE. 

l5^. To find the Middle Latitude. 

Multiply the Departure in miles, by 60, and divide the product by the Difference 
of Longitude in miles ; the quotient vrill be the number of miles in a degree of 
longitude, in the Middle Latitude. To twice the Middle Latitude, add the given 
Latitude, and the sum will be the other Latitude, when one Latitude is North and 
the other South. But from twice the Middle Latitude, subtract the given Latitude, 
and the remainder will be the other Latitude, when the Latitudes are both North 
or both South. 

2d. To find the Distance. 

Add the square of the Difference of Latitude in miles, to the square of the Depart- 
ure in miles, and from the sum of these two squares, extract the square root ; which 
will be the number of miles in the Distance. 

M. To find the Course. 

Multiply the Difference of Latitude in miles, by 60, and divide the product by the 
Distance in miles ; the quotient will be the number of miles in a degree of longi- 
tude, in latitude equal to the Course. 

EXAMPLE.f 

A ship from the Latitude of 3° 24^ South and Longitude 38° 22^ West, sailed 
between the North and East until she made a Departure of 591 miles and arrived in 
North Latitude, and in Longitude 28° 28^ W. ; required the Latitude in, Dis- 
tance sailed, and the Course steered. 

Longitude left 38° 22' W. 

in 28 28 W. 

Difference of Longitude 9 54 = 594 miles. 

\st. To find the Middle Latitude. 

591 X 60 

— — - — ■ = 59.69697 the number of miles in a degree of longitude in the 

^^* Middle Latitude, viz. 5° 46' as per Example XX., page 23. 



The Difference! of the two Latitudes 11 32 

Add the given Latitude, or Latitude left 3 24 S. 

The sum is the other Latitude, or Latitude in 14 56 N., as per Example. Answer. 

Which added to the given Latitude, or Latitude left 3 24 S. 

Gives the Difference of Latitude between the two places 18 20 = 1100 miles. 

2d. To find the Distance. 



i/(1100)2 + (691)2 = 1/1210000 + 349281 = x/l559281 =1248.71174 the number of miles in the Distance 

•' = 1248.712 Answer. 

3c?. To find tlie Course. 

ItOO X 60 
-.oiO'-Ao = 52.85446 the number of miles in a degree of longitude, in latitude equal to the Course, 
■^^*°-' ^^ viz. N. 28° 15' E., as per Example XXI., page 24. Answer. 

* This case cannot be worked by Mercatnr's Sailivg. Bowditch has omitted to give it ! Why ? 

+ Bowditch, in his Examples, never crosses the Equator! Why? 

X In Middle Latitude Sailing, when one Latitude is North and the other South, as in the above Example, 
care must be taken not to mistake the Difference of the two Latitudes, for the Difference of Latitude be- 
tween the two places, which is really the sum of the two Latitudes. See Question X., for exercise. 



MIDDLE LATITUDE SAILING. 



CASE IX.* 



Given the Distance, Departure, and Dififcrence of Longitude, to find both Latitudes 
and the Course. 

RULE. 

l6-^. To find the Difference of Latitude. 

From the square of the Distance in miles, subtract the square of the Departure in 
miles, and from the rcmaiiuler, extract the square root, which will be the number of 
miles in the Diflerence of Latitude. 

2d. To find the Course. 

Multiply the Difference of Latitude in miles, by 60, and divide the product by the 
Distance in miles ; the quotient will be the number of miles in a degree of longitude, 
in latitude equal to the Course. 

2>d. To find the Middle Latitude. 

Multiply the Departure in miles, by 60, and divide the product by the Difference 
of Longitude in miles ; the quotient will be the number of miles in a degree of lon- 
gitude, in the Middle Latitude. 

4ith. To find the two Latitudes. 

If in North Latitude sailing South, or in South Latitude sailing North, to the 
Middle Latitude (which is half the sum of the two latitudes) add half the Difference 
of Latitude in degrees and minutes ; and the sum will be the number of degrees 
and minutes in the Latitude left. And from the Middle Latitude subtract half the 
Difference of Latitude in degrees and minutes ; and the remainder will be the num- 
ber of degrees and minutes in the Latitude in. Again, if in North Latitude sailing 
North, or in South Latitude sailing South, to the ^Iiddle Latitude add half the Dif- 
erence of Latitude in degrees and minutes, and the sum will be the number of 
degrees and minutes in the Latitude in. And from the Middle Latitude subtract 
half the Difference of Latitude in degrees and minutes, and the remainder will be 
the number of degrees and minutes in the Latitude left. 

'^^ When the ship crosses the Equator, subtract twice the Middle Latitude, from 
the Difference of Latitude, in degrees, &c., between the two places, and one-half of 
the remainder will be the less of the two Latitudes. Subtract the less Latitude from 
the Difference of Latitude, in degrees, &c., between the two places, and the remainder 
will be the greater of the two Latitudes, but of a different name from that of the less 
Latitude. 

EXAMPLE. 

A ship in North Latitude, and Longitude 51° 26^ W., sailed between the South 
and East 648 miles, when it was found she had made a Departure of 543 miles, and 
arrived in Longitude 41° 14'' W. ; required the Latitude left, the Latitude in, and 
the Course steered. 

\st. To find the Difference of Latitude. 



|./(64S)2 — (543)2 = |/41990J: — 294849 = |/l25055 = 353.63399 the number of miles in the Difference of 

Li 

2d. To find the Course. 



354 

Latitude = — = 5° 54' Answer. 
60 



648 



353.fi.34 X 60 

■■ 32.74389 the number of miles in a degree of longitude in latitude equal to the Course, viz. 
S. 56° 56' E., as per Example XXII., page 24. Answer. 

Zd. To find the Middle Latitude. 

Longitude left 51° 26' W. 

in 41 14 W. 



Difference of Longitude 10 12 == 612 miles. 

543 X fiO 

— — — — = 53.23529 the number of miles in a degree of longitude in the Middle Latitude, viz. 27° 28', as per 
^^^ Example XXIII., page 24. Answer. 

Ath. To find the two Latitudes. 

Half the sum of the two Latitudes (the Middle Latitude) 27° 28' 27° 28' 

Half the difference of the two Latitudes (the Difference of Latitude) -|- 2 57 — 2 57 

The sumf gives the Latitude left 30 25 N. 24 31 N. 

Answer. The difference! givet^ the Lat- 
itude in. Answer. 

* This Case cannot be worked by Murcatnr's Sailing. Bowditch has omitted to give it ! Why? 
t See Roche's Kuclid, Book V., Prop. XXX. (by the late M.uitin Koche, formerly Professor of Mathematics 
iu the U. S. Navy). 



56 MIDDLE LATITUDE SAILING. 

. QUESTIONS 
To exercise the Learner in the foregoing Rules. 

QUESTION L 

Required the Distance and Course between two places, one in the Latitude of 
37° 55^ North, and the Longitude of 54° 23^ West ; the other in the Latitude of 
32° 38^ North, and the Longitude of 17° 05^ West. 

BY CASE I. 

Latitude of one place 37° 6^ N 57° 55' Longitude of one place 54° 23' W. 

" " the other place...32 38 N 32 38 " " the other place 17 05 W. 

DifiFerence of Latitude 5 17 2)70 33 Difference of Longitude 37 18 

60 60 

35 17 

Dif. of Latitude in milea 317 Middle Latitude. Dif. of Longitude in miles 2238 

1st. To find the Departure. 

Bj the Table of miles in a degree of Longitude in every Latitude, as per Example XV., page 16. 
The Middle Latitude 35° 17' gives 48.97821 the number of miles in a degree of Longitude. 

2238 V 48 97821 

^ — — = 1826.88723 the number of nules in the Departure = 1826.887 Answer. 

60 

2d. To find the Distance. 

l/(317)2 -f (1826.887)2 = -1/100489 + 3337616.110769 = t/ 3438005.1107 69 = 1854.18583 the number of miles 
' ' ' in the Distance =« 

1854 Answer. 

Zd. To find the Course. 

—^ = 10.25782 the number of miles in a degree of Longitude in Latitude equal to the Course, viz.: 

1854.186 g, 80° 09' E., or N. 80° 09' W., as per Example XXIV., page 24. Answer. 

QUESTION XL 

Required the direct Distance and Course from a place in the Latitude of 36° 55'' 
South, and the Longitude of 20° 00^ East, to another place in the Latitude of 32° 38^ 
South, and the Longitude of 8° 54^ West. 

BY CASE I. 

Latitude of one place 36° 65' S 36° 55' Longitude of one place 20° 00' E. 

•' the other place 32 38 S 32 38 " the other place 8 64 W. 

Difference of Latitude 4 17 2)69 33 Difference of Longitude 28 54 

60 60 

34 47 

Dif. of Latitude in miles 257 Middle Latitude, Dif. of Longitude in miles 1734 

1st. To find the Departure. 

By the Table of Miles in a degree of Longitude in every Latitude, as per Example XVI., page 16, 
The Middle Latitude 34° 47' gives 49.27884 the number of miles in a degree of Longitude. 

1734 X 49.27884 ^ ^424.16848 the number of miles in the Departure = 1424.158 Answer. 
60 



2d. To find the Distance. 

1 /(•757VJ 4. a424.168j2 == i/66049 + 2028226.008964 =-i/ 209427 5. 008964 = 1447.16102 the number of milea 
y ^ ^ ^ '^ ' V y in the Distance = 

1447 Answer. 

Zd. To find the Course. 

^^' ^ = 10.65535 the number of miles in a degree of Longitude in Latitude equal to the Course, viz. : 
1447.161 N. 79° 46' W., because the place bound to, is in a less South Latitude than the place 

left, and to the Westward of it, as per Example XXV., page 25. Answer. 

QUESTION IIL 

A ship from the Latitude of 37° 30^ South and the Longitude of 60° 00^ East, sails 
North 79° 56^ West, 202 miles ; required the Latitude and Longitude in. 



MIDDLE LATITUDE SAILING. 57 

BY CASE III. 

IsL Tojind the Difference of Latitude. 

By the Table of Allies in a ilogrce of Longitude in every Latitude, as per Example XYIT., page 16, 
The Course N. 79° 56' W. gives 10.48761 the number of miles in a degiee of longitude in latitude equal to it. 

'^-— '■ — — = 35.30829 the number of miles in the Difference of Latitude = — = 0"= 35' N. Answer 

CO 60 

Latitude left 37° 30' S 37° 30' 

Difference of Latitude 35 N. 



Latitude in 36 55 S 36 55 

A uswer. 

2 )74 -25 

37 13 Middle Latitude. 
Id. Tojind the Departure. 

By the Table of Miles in a degree of Longitude in every Latitude, as per Example XYIIT., page 16, 

The Complement of the Course, viz. 10° 04, gives 59.07621 the number of miles in a degree of longitude in 

latitude equal to it. 

202 X 59.07621 

— -: = 198.88991 the number of miles in the Departure = 198.890. Answer. 

60 

Zd. Tojind the Difference of Longitude. 

By the Table of Miles in a degree of Longitude in every Latitude, as per Example XIX., page 16, 
Tlie Middle Latitude 37° 13' gives 47.78069 the number of miles in a degree of Longitude. 

-1— • • = 249.75358 the number of miles in the Difference of Longitude = -— - = 4° 10' W. 

47.7SU6y ° 60 

Which subtracted from the Longitude left 60 00 E. 



Gives the Longitude in 55 50 E. 

Answer. 

QUESTION IV. 

A ship from the Latitude of 34° 35^ North and the Longitude of 45° 16^ West, 
sails South 83° 36^ East, 101 miles ; required her Latitude and Longitude. 

BY CASE III. 

\st. To find the Difference of Latitude. 

By the Table of miles in a degree of Longitude in every Latitude, as per Example XX., page 16, 
The Course S. 83° 36' E. gives 6.68813 the number of miles in a degree of Longitude in Latitude equal to it. 

'— — '- =^ 11.25835 the number of miles in the Difference of Latitude = = 0° 11' S. Answer. 

Latitude left 34° 35' N 34° 35' 

Difference of Latitude 11 S. 



Latitude in 34 24 N 34 24 

Answer. 

2)68 59 

34 30 Middle Latitude. 

2d. To find the Departure. 

By the Table of miles in a degree of Longitude in every Latitude, as per Example XXI., page 17, 

The Complement of the Course, viz. 6° 24' gives 59.62592 the number of miles in a degree of longitude in 

latitude equal to it. 

101 X 59.62592 

. = 100.37029 the number of miles in the Departure = 100.370. Answer. 

60 

Zd. To find the Differ enoe of Longitude. 

By the Table of miles in a degree of Longitude in every Latitude, on page 9, 

The Middle Latitude 34° 30' gives 49.44761 the number of miles in a degree of Longitude. 

100 370 X 60 122 

'■ = 121.78950 the number of miles in the Difference of Longitude = = 2° 02' E. 

49.44761 ■^ 60 

Which subtracted from the Longitude left 45 16 W. 



Giyes the Longitude in 43 14 W. 

Answer. 



58 MIDDLE LATITUDE SAILING. 

QUESTION V. 

A ship in the Latitude of 49° 57^ North, and the Longitude of 15° W West, sails 
South-Wcsterly till her Departure is 789 miles, and Latitude in 39° 20^ North ; re- 
quired the Distance, Course and Longitude in. 

BY CASE II. 

Latitude left 49° 57' N 49° 57 

in 39 20 N 39 20 



Difference of Latitude 10 37 2)89 17 

60 

44 39 Middle Latitude. 

Dif. of Lat. in miles... 637 



1st. To find the Distance. 



1/(637)2 + (789)2 =i/405769 + 622521 = i/ 1028290 = 1014.04635 the number of miles in the Distance 
^ y * = 1014, Answer. 

2cZ. To find the Course. 

637 V 6C 

-————- == 37.69058 the number of miles in a degree of longitude in latitude equal to the Course, viz., 
1014.04635 g 510 05/ -^y^^ ^s per Example XXVI., page 25. Answer. Because the place bound to, is 

in a less North latitude than the place left and to the Westward of it. 

3^. To find the Difference of Longitude. 

By the Table of Miles in a degree of Longitude in every Latitude, as per Example XXII., page 17, 
The Middle Latitude 44° 39' gives 42.68473 the number of miles in a degree of longitude. 

^ — = 1109.06172 the number of miles in the Difference of Longitude = = 18° 29' W. 

42.68473 60 

Which added to the Longitude left 15 16 W. 

Gives the Longitude in 33 45 W. Answer. 

QUESTION VL 

A ship in the Latitude of 42° 30^ North, and the Longitude of 58° bV West, sails 
S.E. by S. 591 miles ; required the Latitude and Longitude in. 

BY CASE III. 

\st. To find the Difference of Latitude 

By the Table of Points opposite to the Compass, on page 10, 

The Course S.E. by S. 3 points = 33° 45' gives 49.88810 the number of miles in a degree of longitude in 

latitude equal to it. 

— '■ = 491.39778 the number of miles in the Difference of Latitude = ■ — =8° 11' S. Answer. 

60 60 

Latitude left 42° 30' N 42° 30' 

Difference of Latitude 8 11 S. 



Latitude in 34 19 N 34 19 

Answer. 

2)76 49 



38 25 Middle Latitude. 

2d. To find the Departtire. 

By the Table of Points opposite to the Compass, on page 10, 

The Complement of the Course, viz., 5 points = 56° 15' gives 33.33418 the number of miles in a degree of 

longitude in latitude equal to it. 

591 V^ 33 33418 

. '' ' = 328.34167 the number of miles in the Departure = 328.342 Answer. 

60 

Sd. To find the Difference of Longitude. 

By the Table of Miles in a degree of Longitude in every Latitude, as per Example XXIII., page 17, 
The Middle Latitude 38° 25' gives 47.01065 the number of miles in a degree of longitude. 

328 342 V 60 419 

419.06504 the number of miles in the Difference of Longitude = — -- = 6° 59' E. 



47.01065 ■ 60 

Which subtracted from the Longitude left 58 51 W. 



Gives the Longitude in 51 52 W. 

Answer. 



MIDDLE LATITUDE SAILING. 59 

QUESTION VIL 

Suppose a ship sailing; from a pliiee in the Latitude of 49° 57^ North, and the 
Longitude of 30° 00^ West, makes a Course of S. 3*.)° W., and then, by observation, 
is iu the Latitude of 45° 31^ A'orth; required the Distance and Longitude in. 

BY CASE IV. 

Latitude left 49057' N 49° 57' 

in 45 31 N 45 31 



Difiference of Latitude 4 26 2;9o 28 

60 

47 44 Middle Latitude. 

Dif. of Lat. in miles... 266 

IsL Tojind the Distance. 

By the Table of Miles in a degree of Longitude in every Latitude, on page 9, 
The Course S. 39° W. gives 46.62882 the number of miles in a degi'ee of longitude in latitude equal to it. 

'- ^^ = 342.27758 the number of miles in the Distance = 342.3 Answer. 

46.62882 

2d. Tojind the Departure. 

By the Table of Miles in a degree of Longitude in every Latitude, on page 9, 

The Complement of the Course, viz., 51° gives 37.75922 the number of miles in a degree of longitude in lati- 
tude equal to it. 
Q 1.7 o'Tfi y 37 75Q22 

-^^ — '—^ — = 215.40251 the number of miles in the Departure = 215.403 Answer. 

60 

2>d. Tojind the Difference oj Longitude. 

By the Table of Miles in a degree of Longitude in every Latitude, as per Example XXIY., page 17, 
The Middle Latitude 47° 44' gives 40.35487 the number of miles in a degree of Longitude. 

" ^ 40 V 48 ^ = 320.26320 the number of miles in the Difference of Longitude = -^ = 5° 20' W. 

Which added to the Longitude left 30 00 W. 
Gives the Longitude in 35 20 W. Answer. 

QUESTION VIIL 

A ship in the Latitude of 50° 10^ South and the Longitude of 30° 00^ East, 
sails E.S.E. until her Departure is 957 miles. Required her Distance sailed, and 
Latitude and Longitude in. 

BY CASE VI. 

1st. To jind the Distance. 

By the Table of Points opposite to the Compass, on page 10, 

The Complement of the Course, viz. 2 points = 22° 30' gives 55.43268 the number of miles in a degree of lon- 
gitude in latitude equal to it. 

957 X 60 

,, ,„, ■ = 1035.85105 the number of miles in the Distance = 1036. Answer. 
00.43268 

2d. Tojind the Difference oj Latitude. 

By the Table of Points opposite to the Compass, on page 10, 

The Course E.S.E. 6 points = 67° 30' gives 22.96101 the number of miles in a degree of longitude in latitude 

equal to it. 

1035.851 X 22.96101 396 

-— = 396.40309 the number of miles in the Difference of Latitude = -— = 6° 36' S. 

^ 60 Answer. 

Latitude left 80° 10' S 50° 10' 

Difference of Latitude 6 36 S. 



Latitude in 56 46 S 56 46 

Answer. 

2)106 56 

53 28 Middle Latitude. 

2>d. Tojind the Difference oj Longitude. 

By the Table of Miles in a degree of Longitude in every Latitude, as per Example XXV., page 17, 
The Middle Latitude 53° 28' gives 35.71737 the number of miles in a degree of Longitude. 

957 X 60 1608 

-^nrr^^ = 1607.62117 the number of miles in the Difference of Longitude = — — = 26° 48' E. 
oo.ilVoT 60 

Which added to the Longitude left 30 00 E. 



Gives the Longitude in 56 48 E. 

Answer. 



60 MIDDLE LATITUDE SAILING. 



QUESTION IX. 

A ship in the Latitude of 49° 30^ North, and the Longitude of 25° 00^ West, sails 
South-Easterly 645 miles, until her Departure from the meridian is 500 miles ; 
required the Course, and the Latitude and Longitude in. 

BY CASE VII. 

IsL Tojind the Course. 

fiOO X 60 

■ — = 46.51163 the number of miles in a degree of longitude, in latitude equal to the Complement 

**** of the Course, viz., 39° 11', as per Example XXVII., page 25. 

The Complement subtracted from 90° gives the Course S. 50° 49' E., as per Question Answer. 

2d. To Jind the Difference of Latitude. 

By the Table of Miles in a degree of Longitude in every Latitude, as per Example XXVI., page 17, 

The Course S. 50° 49' E. gives 37.90817 the number of miles in a degree of longitude in latitude equal to it. 

645 X 37.90817 408 

— = 407.51283 the number of miles in the Difference of Latitude = -— - = 6° 48' S. Answer. 

dO 60 

Latitude left 49° 30' N 49° 3^ 

Difference of Latitude 6 48 S. 



Latitudein 42 42 N 42 42 

Answer. 

2)92 12 

46 06 Middle Latitude. 

Zd. To find the Difference of Longitude. 

By the Table of miles in a degree of Longitude in every Latitude, as per Example XXVII., page 18, 

The Middle Latitude 46° 06^ gives 41.60399 the number of miles in a degree of longitude. 

500 X 60 721 

= 721.08468 the number of miles in the Difference of Longitude = — =12° 01' E. 

41.60399 *' 60 

Which subtracted from the Longitude left 25 00 W. 



Gives the Longitude in 12 59 "W. 

Answer. 



QUESTION X.^ 



A ship in the Latitude of 3° 24^ South and Longitude 38° 22^ West, sailed 
between the South and East until she made a Departure of 591 milee, and arrived 
in Longitude 28° 28^ West. Required the Latitude in, Distance sailed, and Course 
steered. 



BY CASE VIII. 



Longitude left 38° 22' W. 

in 28 28 W. 



Difference of Longitude 9 54 = 594 miles. 

\st. To find the Middle Latitude. 

591 X 60 

— — • = 59.69697 the number of miles in a degree of longitude in the 

''^^ Middle Latitude, viz. 5° 46' as per Example XX., page 23. 



The sum of the two Latitudes 11 32 

Subtract the given Latitude, or Latitude left 3 24 S. 



And the remainder will be the other Latitude, or Latitude in 8 08 S., as per Example. Answer. 
From which subtract the given Latitude, or Latitude left 3 24 S. 



Gives the Difference of Latitude between the two places 4 44 = 284 miles. 

2(7. To find the Distance. 



-|/(284)2 -f (591)2 = -i/'80656 + 349281 = -i/429937 = 655.69581 the number of miles in the Distance = 
^ *^ *^ 655.696 Answer. 

2>d. To find the Course. 

284 X 60 

-—-——- "= 25.98765 the number of miles in a degree of longitude in latitude equal to the Course, viz. 

655.69b s. 640 20' E., as per Example XXXI., page 26. Answer. 

* This Question is not in Bowditch. See Example of Case yill. 



MIDDLE LATITUDE SAILING. 



61 



COMPOUND COURSES. 

Given the latitude and longitude of the place sailed from, and the different 
courses and distances sailed, to lind the difference of latitude made good, the depart- 
ure made good, the latitude in, the distance made good, the course made good, the 
difference of longitude made good, and the longitude in. 

RULE. 

Isi. Tojind the Difference of Latitude made good, and the Departure made good. 

Make a Traverse Table of the different Courses and Distances sailed, and by it 
find the Difference of Latitude made good, and the Departure made good, by Part II. 
of the Rule for Traverse Sailing, on page 35. 

2(7. Tojind the Distance made good. 
Add the square of the Difference of Latitude in miles made good, to the square of 
the Departure in miles made good, and from the sum of these two squares extract 
the square root, vrhich will be the Distance in miles made good, by Part I. of the 
Rule for Case VI. Plane Sailing, on page 31. 

3cZ. Tojind the Course made good. 
Multiply the Difference of Latitude in miles made good, by 60 and divide the pro- 
duct by the Distance in miles made good, the quotient will be the number of miles 
in a degree of longitude, in latitude equal to the Course made good, by Part I. of 
the Rule for Case V., Middle Latitude Sailing, on page 51. 

4/7i. To Jind the Difference of Longitude made good. 

Multiply the Departure in miles made good, by 60 and divide the product by the 
number of miles in a degree of longitude in the Middle Latitude, the quotient will 
be the number of miles in the Difference of Longitude made good, by Part III. of 
the Rule for Case V., Middle Latitude Sailing, on page 51. 

EXAMPLE. 

A ship takes her departure from Cape Henlopen, in the Latitude of 38° 47^ North 
and the Longitude of 75° 05^ West, bearing W, by N., distant 20 miles, and then 
sails on the following true courses, viz., E.N.E; 15 miles, S.E. 26 miles South 16 
miles, W.S.W. 6 miles, N.W. 10 miles, and East 30 miles ; required the Difference 
of Latitude made good, the Departure made good, the Latitude in, the Distance 
made good, the Course made good, the Difference of Longitude made good, and the 
Longitude in. 

\st. To find the Difference of Latitude made good, and the Departure made good 
Having calculated all the Differences of Latitude and Departures made on the 
several Courses and Distances sailed, by Case I., Plane Sailing, arrange them, ac- 
cording to the Rule, in the form of a 

TRAVERSE TABLE. 







Difference of Latitude. 


Departure. 


Course. 


Distance. 


N. 


S. 


E. W. 


E. by S. 


20 




3.9 


19.6 


E.N.E. 


15 


5.7 




13.9 




S.E. 


26 




18.4 


18.4 




South 


16 




16.0 






• W.S.W. 


6 




2.3 




5.5 


1 N.W. 


10 


7.1 






7.1 


East 


30 






30.0 








12.8 


40.6 

12.8 


81.9 
12.6 


12.6 

1 


Remain 


der 27.8 


69.3 Rer 


aainder 1 




Differ< 


mce of Latitude made good. 


Departure made good. 



62 MIDDLE LATITUDE SAILING. 

2il. To find the Distavce made good. 



-i/(-27.S)''2 + (69.;i)2 = W-,1±U + 4802.49 =1/ 5575.33 = 74.60813 the nunibor of miles iu the Distance = 
*'*'*' 74.7 Answer. 

3cZ. To find the Course made good. 

27 8 V 60 

-—- • = 22.33885 the numljer of milos in a dojrroe of lonRitndo in latitude equal to the Pourso ma^lo 

'*-66813 goofj^ vjy_^ g_ 680 OS/ -y .^^ p^.p Example XXTITI., p. 25. Because the DifTerenc- of \mX\- 

tude made good is to the Southward, and the Departure made good is to the Eastward. 

Latitude of Cape Henlopen 38° 47' N 38° 47 

Difference of Latitude 28 S. 

Latitude in 38 19 N 38 19 

Answer. 

2)77 06 

38 33 Middle Latitude. 

Ath. To find the Difference of Longitude made good. 

By the Table of Miles in a degree of Longitude in every Latitude, as per Example XXYIIT., page 18, 

The Middle Latitude 38° 33' gives 46.92375 the number of miles in a degree of longitude. 

AQ ^ \/ fin 8Q 

— = 88.61184 the number of miles in the Difference of Longitude made good = — ~ = 1° 29' E 
46.92375 60 

Which subtracted from the Longitude of Cape Henlopen 75 05 W 

Gives, the Longitude in 73 36 W. 

Answer. 

Remark. — This example is the same as that given by Bowditch, at the end of 
Mercator's Sailing, in his Practical Navigator, where he has attempted . to exem- 
plify the working of compound courses, both by Middle Latitude and Mercator's 
Sailing. 



MERC A TOR'S SAILING. 



Mercator's Sailixg is called after Gerard Mercator. a Flemisli geographer, 
who invented and published in the year 156G his chart, on which the meridians are 
parallel straiglit lines, and consequently all the parallels of latitude are equal to the 
Equator. The degrees of latitude and the meridian distances are increased in the 
same proportion. Hence it follows that the Rhumbs which form equal angles with 
the meridians will be straight linos, which renders this projection of the Earth's 
surface much more easy and proper for the mariner's use than any other. 

Mercator's Chart m.ay be roughly explained as follows : — Suppose the globe to be 
surrounded by a hollow cylinder which touches it everywhere at the Equator, and 
that the degrees of latitude and longitude are expanded proportionally from the 
Equator towards the Poles, until it meets the concave of the cylinder, marking upoa 
it all the lines that are drawn on its surface. Then the cylinder being opened its 
whole length, and unrolled, will be Mercator's Chart, on which the places bear the 
same relation and proportion to each other, as do the corresp.mding places on the 
surface of the globe. 

The increased meridian between the two parallels is called the meridional differ- 
ence of latitude, in contradistinction to the actual difference, which is called the 
proper difference of latitude. The meridional difference of latitude is obtained by 
finding in Table III. of the Epitome the meridional parts of the two latitudes, and 
taking the difference between them, when the latitudes are of the same name, but 
by adding them together when the latitudes are of different names. 

In Table III. of the Epitome, the meridional parts are only calculated to the 
nearest mile or minute, and therefore Mercator's Sailing is not strictly accurate 
when the difference of latitude is small, consequently it is not much used on ship- 
board in calculating short runs and day's work. 



To find the Meridional Difference of Latitude. 

EXAMPLE I. 

What is the ^Meridional Difference of Latitude, between Cape Cod Light, and the 
Island of St. Mary, one of the Azores (or Western Islands) ? 

Latitude of Cape Cod light... 42° 0-3' N Meridional Parts 2786 as per Table III., of the Epitome, 

" St. Mary 36 59 N " " 2391" " " " " 

Proper Difference of Latitude 5 04 395 

60 Meridional Difference of La^titude in miles. 

Answer. 

Proper Dif. of Lat. in miles... 304 

EXAMPLE IL 

What is the Meridional Difference of Latitude, between two places, one in W^ 56'' 
North Latitude, and the other in 3° 24^ South Latitude. 

Latitude of one place 14° 56' N Meridional Parts 906, as per Table IIT., of the Epitome, 

" the other place... 3 24 S " " 204" " " " " 

Proper Difference of Latitude 18 20 1110 

60 Meridional Difference of Latitude in miles. 

Answer, 

Proper Dif of T^t. in miles.... 1100 



(03) 



64 MERCATOll'S SAILING. 

CASE I. 

Given the Latitudes and Longitudes of two places, to find the Course (or their 
Bearing) and Distance. 

RULE. 

isi. To find a Distance correspnndinr/ to the Meridional Difference of Latitiide and the 

Difference of Longitude. 

Add the square of the Meridional Difference of Latitude in miles, to the square of 
the Difference of Longitude in miles, and from the sum of these two squares extract 
the square root; which will be a Distance in miles (but not the dis^a,nce between 
the two places). 

2d. To find the Course from this Distance. 

Multiply the Meridional Difference of Latitude in miles, by 60, and divide the 
product by the distance obtained above, the quotient will be the number of miles in 
a degree of longitude in latitude equal to the Course. 

Zd. To find the Distance between the two 'places.* 

Multiply the Proper Difference of Latitude in miles, by 60, and divide the product 
by the miles in a degree of longitude in latitude equal to the Course ; the quotient 
will be the number of miles in the Distance between the two places. 

EXAMPLE. 

Required the Course and Distance between Cape Cod lighthouse, in the Latitude 
of 42° 03^ North, and the Longitude of 70° 04^ West ; and the Island of St. Mary 
(one of the Azores or Western Islands), in the Latitude of 36° 59^ North, and the 
Longitude of 25° 10^ West. 

8i^^ See our remark on this question under Case I., Middle Latitude Sailing. 

Cape Cod's Latitude 42° 03' N Meridional Partsf 2786 Cape Cod's Longitude.... 70*^ 04' W. 

St. Mary's " 36 59 N " " 2391 St. Mary's " 25 10 W. 

Proper Dif. of Latitude... 5 04 Meridional Dif. of Latitude 395 Difference of Longitude 44 54 

60 60 



Proper Dif. of Lat. in miles 304 Dif. of Longitude in miles 2694 

\st. To find a Distance corresponding to the Meridional Difference of Latitude and 
the Difference of Longitude. 



t/(395)2 -I- (2694)2 = i/l66025 + 7257636 = -1/7413661 = 2722.80389 the number of miles in a Distance 

corresponding to the Meridional Dif. 
of Lat. and the Dif of Long. 

2d. To find the Course. 

395 ^ 60 

■ = 8.70430 the number of miles in a degree of longitude, in latitude equal to the Course, 

2722.804 yiz.^ g. gio 40/ e., as per Example XXIX., page 26. Because the Island of St. Mary is 

in a less North latitude than Cape Cod, therefore St. Mary is to the Soiithward of Cape 
Cod; and because the Island of St. Mary is in a less West longitude than Cape Cod, 
therefore St. Mary is to the Eastward of Cape Cod. Hence Cape Cod lighthouse bears 
from the Island of St. Mary, on the point of the compass directly opposite, viz., 
N. 81° 40' W. Answer. 

Zd. To find the Distance. 

^04- V 60 

Z> — == 2095.51601 the number of miles in the Distance = 2096.t Answer. 

8.70430 * 

* The Distance may be found otherwise, by the Departure, thus : 

As Mer. Dif. of Lat. : Dif. of Long. : : Proper Dif. of Lat. : Departure. 
395 : 2694 : : 304 : 2073.35696 



1 /(Prop. Dif. of Lat.)2 + (Departure)^ =-|/ (304)2 + (2073.35696)2 == i/ 4391225.0835804416 = 2095.52501 = 
*' *^ *^ 2096. Answer. 

+ The Meridional Parts are taken from Table TIT., of the Epitome. 

\ Bowditch's answer is not strictly correct. His course by logarithms proportioned to seconds is 81° 39' 31", 
the secant of which put in the 26. canon, gives 3.32129, the exact logarithm answering to the Distance 
2095.524, which at most is only = 2096 and not 2098 miles, as given by him. 



MERCATOR'S SAILING. 65 

CASE II. 

Given both Latitudes and the Departure to find the Distance, Course, and Differ- 
ence of Longitude. 

RULE. 
1st. To find the Distance. 

Add the square of the Proper Difference of Latitude in miles, to the squnro of the 
Departure in miles, and from the sum of these two squares, extract the square root ; 
which will be the number of miles in the Distance. 

2d. To find the Course. 

Multiply the Proper Difference of Latitude in miles, by 60, and divide the product 
by the Distance in miles ; the quotient will be the number of miles in a degree of 
longitude, in latitude equal to the Course. 

M. To find the Difference of Longitude.* 

Multiply the Meridional Difference of Latitude in miles, by 60, and divide the 
product by the number of miles in a degree of longitude, in Latitude eqiinl to the 
Course; the quotient will be a Distance in miles, corresponding to the Meridional 
Difference of Latitude. Multiply this Distance in miles, by the miles in a Je<^ree 
of longitude, in latitude equal to the Complement of the Course, and divide the pro- 
duct by 60 ; the quotient will be the number of miles in the Difference of Lon- 
gitude. 

EXAMPLE. 

A ship in the Latitude of 49° 57^ North and the Longitude of 15° 16^ West, 
sails South- Westerly until her Departure is 197 miles, and then by observation is in 
the Latitude of 47° 18^ North; required her Distance, Course, and Longitude in. 

Latitude left 49° 57' N Meridional Parts S470 as per Table III., of tlio Kpitnme, 

in 47 18 N " " 3229" " " « " 

Proper Difference of Latitude 2 39 241 

60 Meridional Difiference of Latitude. 

Proper Dif. of Lat. in miles... 159 

Ist. To find the Distance. 

1/(159)2 + (197)2 ^t/25281 + 38809 = t/ 64090 = 253.16003 the number of miles in the Distance == 

253.2 Answer. 

2d. To find the Course. 

159 X 60 

- ocQiftA " == 37.68368 the number of miles in a degree of longitude in latitude equal to the Course, viz.: 
'^'*-^*^ S. 51° 06' W., as per Example XXX., page 26. Answer. 



241 X 



Zd. To find the Difference of Longitude, 

= 383.72049 the number of miles in a Distance corresponding to the Meridional Dif. of Latitu'le. 

By the Table of miles in a degree of Longitude in every Latitude, as per Example XXIX., page !«. 

The Complement of the Course, viz. 38° 54', gives 46.69456 the number of miles in a degree of longitude in 

latitude equal to it. 

'■ ^ — '■ = 298.62766 the number of miles in the Dif. of Longitude = 298.6 = — = 4° 59' W. 

Which added to the Longitude left 15 Ifi W. 



Gives the Longitude in .'20 15 W. 

Answer. 



* The Difference of Longitude may be found otherwise, by the Departure, thus : 

As Proper Dif. of Lat. : Departure : : Meridional Dif. of Lat. : Dif. of Longitude. 

159 : 197 : : 241 : 298.59748 = 299 miles. 

Answer. 



66 MERCATOR'S SAILING. 

CASE III. 

Given one Latitude, the Course and Distance, to find the Difference of Latitude 
and Difference of Longitude. 

RULE. 

l5^. To find the Proper Difference of Latitude. 

Multiply the Distance in miles, by the miles in a degree of longitude, in latitude 
equal to the Course, and divide the product by GO ; the quotient will be the number 
of miles in the Proper Difference of Latitude. 

2d. To find the Difference of Longitude.* 

Multiply the Meridional Difference of Latitude in miles, by 60, and divide the pro- 
duct by the number of miles in a degree of longitude, in latitude equal to the Course ; 
the quotient will be a Distance in miles, corresponding to the Meridional Difference 
of Latitude. Multiply this Distance in miles, by the miles in a degree of longitude, 
in latitude equal to the Complement of the Course, and divide the |)roduct by 60 ; 
the quotient will be the number of miles in the Difference of Longitude. 

EXAMPLE. 

A ship in the Latitude of 42° 30^ North, and the Longitude of 58° 51^ West, sails 
S.W. by S., 300 miles; required the Latitude and Longitude in. 

\st. To find the Proper Difference of Latitude. 

By the Table of Points opposite to the Compass, on page 10, 

The Coxirse S.W. by S. 3 points = 33° 45' gives 49.88810 the number of miles in a degree of longitude in 

latitude equal to it. 

800X49.8S810 249 
— ■ = 249.44050 the number of miles in the Proper Dif. of Lat. = — - = 4° 09' S., as per Example. 

Latitude left 42° 30^ N .Meridional Parts 2822, as per Table III., of the Epitome, 

Proper Difference of Latitude 4 09 S. 

Latitude in 38 21 N " " 2495" " " " " " 

Meridional Difference of Latitude 327 

2d. To find the Difference of Longitude. 

327 X 60 

• -= 393.28016 the number of miles in a Distance corresponding to the Meridional Dif. of Latitude. 

49.^ool0 

By the Table of Points opposite to the Compass, on page 10, 

Tlie Complement of the Course, viz, 5 Points = 56° 15' gives 33.33418 the number of miles in a degree of 

longitude in latitude equal to it. 

393.28016 X 33.33418 218 

— = 2l8.49453t the number of miles in the Dif of Long.= — =3° 38' W. as per Example. 

Which added to the Longitude left 58 61 W. 

Gives the Longitude In 62 29 W. 

Answer. 

* The Difference of Longitude may be found otherwise, by the Departure, thus : 



l/(Distance)2— (Prop. Dif. of Lat.)2 = -|/(300)2 _ (249.4405)2 = -i/goooO — 62220.56304025 == 

-|/27779.43695975 = 166.67164 Departure. 

As Proper Dif. of Lat. : Departure : : Meridional Dif of Lat. : Dif of Longitude. 

249.4405 : 166.67164 : : 327 : 218.49550 = 218 mUes. 

Answer. 

f Eowdi roll's answer is not strictly correct, for by proportion his logarithm 2.33944 answers to 218.495 = 
218 only, and not 219 miles, as given by him. 



MERCATOR'S SAILING. 67 

CASE IV. 

Given both Latitudes and the Course, to find the Distance, and Difference of Lon- 
gitude. 

RULE. 

1st. To find the Distance. 

Multiply the Proper Difference of Latitude in miles, by 60, and divide the product 
by the miles in a degree of longitude, in latitude equal to the Course ; the qliotieut 
will be the number of miles in the Distance. 

2d. To find the Difference of Longitude.* 

Multiply the Meridional Difference of Latitude in miles, by 60, and divide the 
product by the number of miles in a degree of longitude in latitude equal to the 
Course ; the quotient vrill be a Distance in miles, corresponding to the Meridional 
Difference of Latitude. Multiply this Distance in miles, by the miles in a degree of 
longitude, in latitude equal to the Complement of the Course, and divide the pro- 
duct by 60 ; the quotient will be the number of miles in the Difference of Longitude. 

EXAMPLE. 

A ship from the Latitude of 49° 57^ North and the Longitude of 30° We'st, sails 
South 39° West, till she arrives in the Latitude of 47° 44^ North ; required the Dis- 
tance run, and the Longitude in. 

Latitude left 49° 57' N Meridional Parts 3470, as per Table in. of the Epitome. 

in 47 44 " " 3268 " " 



Proper Dif. of Latitude 2 13 Merid. Dif. of Lat. 202 

60 

Prop. Dif. of Lat. in miles... 133 

1st. To find the Distance, 

By the Table of Miles in a degree of Longitude in every Latitude, en page 9, 
The Course S. 39° W. gives 46t62882 the number of miles in a degree of longitude in latitude equal to it. 

-— — — r = 171.13879 the number of miles in the Distance = 171.1 Answer. 
4o,o.^8b2 

2d. To find the Difference of Longitude, 

oryo v/ AA 

— r~— ,„- -- = 259.92508 the number of miles in a Distance corresponding to the Meridional Dif. of Latitude. 
46.62882 

By the Table of Miles in a degree of Longitude in every Latitude, on page 9, 

The Complement of the Course, viz., 51°, gives 37.75922 the number of miles in a degree of longitude in lati- 
tude equal to it. 

S'iQ Q2'i08 V ^7 7'iQ22 164 

I—: ;^ = 163.57614 the number of miles in the Dif. of Long. = 163.6 = — - = 2° 44' W., per Ex. 

60 60 

"Which added to the Longitude left 30 00 W. 



Gives the Longitude in 32 44 W. 

Answer. 



* The Difference of Longitude may be found otherwise, by the Departure, thus : 



i^/(Di8tance)2 _ (Prop. Dif. of Lat.)3 = |/(171.13879)a — (133)2 = 1/29288.4854426641 — 17689 = 

-j/ll599.4854426641 = 107.70091 Departure. 

As Proper Dif. of Lat. : Departure : : Meridional Dif. of Lat. : Dif. of Longitude. 

133 : 107.70091 :: 202 163.57582 = 164 miles. 

Answer, 



68 MERCATOR'S SAILING. 

CASE V. 

Givon both Latitudes and the Distance, to find the Course, and Difference of Lon- 
gitude. 

RULE. 

1st. To find the Course. 

Multiply the Proper Difference of Latitude in miles, by 60, and divide the pro^ 
duct by the Distance in miles ; the quotient will be the number of miles in a degree 
of longitude, in latitude equal to the Course. 

2d. To find the Difference of Longitude.* 

Multiply the Meridional Difference of Latitude in miles, by 60, and divide the pro- 
duct l)y the number of miles in a degree of longitude, in latitude equal to the 
Cfuirse ; the quotient will be a Distance in miles, corresponding to the Meridional 
Difference of Latitude. Multiply this Distance in miles, by the miles in a degree of 
longitude, in latitude equal to the Complement of the Course, and divide the product 
by 60 ; the quotient will be the number of miles in the Difference of Longitude. 

EXAMPLE. 

A slup from the Latitude of 37° North and the Longitude of 32° 16^ West, sails 
300 miles North-Westerly, until she is in the Latitude of 41° North ; required the 
Course steered and Longitude in. 

Latitude in 41° N Meridional Parts 2702, as per Table III. of the Epitome. 

left 37 N " » 2393 " " " " 

Proper Dif. of Latitude 4 Merid. Dif. of Lat. 309 

60 

Prop. Dif. of Lat. in miles... 240 

\st. To find the Course. 

- — = 48.00000 the number of miles in a degree of longitude in latitude equal to the Course, viz., 

300 N. 36° 52' W., as per Example "XVni., page 23, 

2d. To find the Difference of Longitude. 

309 V fiO 
_ = 386.25000 the number of miles in a Distance corresponding to the Meridional Dif of Latitude. 

48.00000 

By the Table of Miles in a degree of Longitude in every Latitude, as per Example XI., page 15, 

The Complement of the Course, viz. 53° 08', gives 35.99741 the number of miles in a degi-ee of longitude in 

latitude equal to it. 

■ ' ^ -^''^"'^" ^^J = 231.73333 the number of miles in the Dif. of Long. = ^ = 3° 52' W., as per Example. 

(jO 60 

Which added to the Longitude left 32 16 W. 

Gives the Longitude in 36 08 W. Answer. 



* The Difference of Longitude may be found otherwise, by the Departure, thus : 



•i/(Di<tance)2 — (Prop. Dif. of Lat.)2 = i/(300)2 _ (240)2 = 1/900OO — 57600 = y^32400 = 180 Departure. 

As Proper Dif. of Lat. : Departure : : Meridional Dif. of Lat. : Dif. of Longitude. 

240 180 : : 309 : 231.75000 = 232 miles. 

Answer. 



MERCATOR"S SAILING. 69 

CASE VI. 

Given one Latitude, the Course and Departure, to find the Distance, Difference of 
Latitude, and Diliereuce of Longitude. 

RULE. 

1st. Toji/id the Distance. 

Multiply the Departure in miles, by 60, and divide the product by the miles in a 
degree of longitude, in latitude equal to the Complement of the Course ; the quo- 
tient will be the number of miles in the Distance. 

2c?. To find the Proper Difference of Latitude^ 

Multiply the Distance in miles, by the miles in a degree of longitude, in latitude 
equal to the Course, and divide the product by 60 ; the quotient will be the number 
of miles in the Proper Difference of Latitude. 



2>d. To find the Difference of Longitude. 



Multiply the Meridional Difference of Latitude in miles, by 60, and divide the 
product by the number of miles in a degree of longitude, in latitude equal to the 
Course ; the quotient will be a Distance in miles, corresponding to the Meridional 
Difference of Latitude. Multiply this Distance in miles, by the miles in a degree of 
longitude, in latitude equal to the Complement of the Course, and divide the pro- 
duct by 60 ; the quotient will be the number of miles in the Difference of Longitude. 

EXAMPLE. 

A ship from the Latitude of 50° 10^ South and the Longitude of 30° East, sails 
E.S.E. until her Departure is 160 miles ; required the Distance sailed, and the Lati- 
tude and Longitude in. 

\st. To find the Distance. 

By the Table of Points opposite to the Compass, on page 10, 

The Complement of the Course, viz., 2 Points = 22° 30', gives 55.43268 the nnmber of miles in a degree of 

longitude in latitude equal to it. 

/^. _ = 173.18304 the nnmber of miles in the Distance = 173.2 Answer. 

55.43268 

2d. To find the Proper Difference of Latitude. 

By the Table of Points opposite to the Compass, on page 10, 

The Course E.S.E. 6 points = 67" 30^ gives 22.96101 the number of miles in a degree of longitude in latitude 

equal to it. 

173.18304 X 22.96101 66 

. .== 66.27429 the number of miles in the Pi-oper Dif. of Lat. = 66.3 = — = 1° 6' S., as per Ex. 

Latitude left 50° W S Meridional Parts 3490, as per Table III. of the Epitome. 

Proper Dif. of latitude 1 06 S. 

Latitude in 51 16 S " " 3594 " " « « 

Answer. 

Meridional Dif. of Lat. 104 

ScZ. To find the Difference of Longitude. 

104 X 60 

■ - = 271.76505 the number of miles in a DLstance corresponding to the Meridional Dif of Latitude. 

271.76505X55.4.3268 „.,„„._ ,^ u . -, - , t... . x 251 

. — = 2ol.O/ i 1 the number of miles m the Dif of Long. = — =4° 11' E., as per Example. 

dO 60 

Which added to the Longitude left 30 00 E. 

Gives the Longitude in 34 11 B. 

A nswer, 

• The Difference of Longitude may be found otherwise, by the Departure, thus : 

As Proper Dif. of Lat. : Departure : : Meridional Dif. of Lat. : Dif. of Longitude. 

66.27429 : 160 :: 104 : 251.07776 =2.51 miles. 

Answer. 



70 MERCArOR'S SAILING. 

CASE VII. 

Given one Latitude, the Distance sailed, and Departure, to find the Course, Differ- 
ence of Latitude, and Difference of Longitude. 

RULE. 

1st. Thjind the Course. 

Multiply the Departure in miles, by 6<3, and divide the product by the Distance 
in miles ; the quotient vf'iW be the number of miles in a degree of longitude, in lati- 
tu<lo equal to the Complement of the Course. Subtract the Complement from 90° 
and the remainder will be the Course in degrees, &c. 

2c?. To jind the Proper Difference of Latitude.* 

Multiply the Distance in miles, by the miles in a degree of longitude, in latitude 
equal tu the Course, and divide the product by 60 ; the quotient will be the number 
of miles in the Proper Difference of Latitude. 

?>d. To find the Difference of Longitude.^ 

Multiply the Meridional Difference of Latitude in miles, by 60, and divide the pro- 
duct by the number of miles in a degree of longitude, in latitude equal to the 
Course ; the quotient vrill be a Distance in miles, corresponding to the Meridional 
Difference of Latitude. Multiply this Distance in miles, by the miles in a degree of 
longitude, in latitude equal to the Complement of the Course, and divide the product 
by 60 ; the quotient vrill be the number of miles in the Difference of Longitude. 

EXAMPLE. 

A ship in the Latitude of 49° 30^ North and the Longitude of 25° West, sails 
South-Easterly 215 miles, making 167 miles Departure ; required the Course steered, 
and the Latitude and Longitude in. 

\st. To find the Course, 

— -— — ^ 46.60465 the mimber of miles in a degree of longitude in latitude equal to the Complement 
'-^^ of the Course, viz., 39° 02', as per Example XIX., page 2-3. Answer. 

The Complement subtracted from 90° 00', gives the Coiarse S. 60° 58' E., as per Example. 

2d. To find the Proper Difference of Latitude. 

By the Table of Miles in a degree of Longitude in every Latitude, as per Example XIII., page 15, 
The Course, S. 50° 58' E., gives 37.78631 the number of miles in a degree of longitude in latitude equal to it. 
215 X 3<.<S6ol ^ j35^oQ9^ ^j number of miles in the Proper Dif. of Lat. = 135.4 = — - = 2° 15' S., per Ex. 

Liititude left 49° 30' N Meridional Parts 3428, as per Table III. of the Epitome. 

rnipor Dif. of Latitude 2 15 S. 



1-atitudein 47 15 N " " 3225 " « « « 

Answer. 

Merid. Dif. ofLat. 203 

3cZ. To find the Difference of Longitude. 

203 ^ 60 

= 322.33896 the number of miles in a Distance corresponding to the Meridional Dif. of Latitude. 

37.786;il 

'■ — ^ — ^ — '■ = 250.37491 the number of miles in the Dif. of Long. = 250.4 : 

V^'hich subtracted from the Longitude left 25 00 "W. 



Gives the Longitude in 20 50 W. 

Answer. 



• The Proper Difference of Latitude may be found otherwise, by Case V. of Plane Sailing, thus : 



j/(Distance)a — (Departure)^ = |/^(215)2 _ (167)2 = |/46225 — 27889 = -|/l8336 = 135.41049 =» 

135.4 Prop. Dif. OfLat. 
-f- The DifTerence of Longitude may be found otherwise, by the Departure, thus: 

As Proper Dif. of Lat. : Departure : : Meridional Dif. of Lat. : Dif. of Longitude. 

135.41049 : 167 : : 203 • 250.35727 = 250.4 miles. 

Answer. 



MERCATOR'S SAILING. 71 

CASE VIIL* 

Given one Latitude, the Course, and Difference of Longitude, to find the Differ- 
ence of Latitude and Distance. 

RULE. 

1st. Tojind the Distance corresponding to the Meridional Difference of Latitude. 

Multiply the Difference of Longitude in miles, by 60, and divide the product l)y 
the miles in a degree of longitude, in latitude equal to the Complement of the Course ; 
the quotient vrill be the number of miles in a Distance, corresponding to the Meri- 
dional Difference of Latitude. 

2d. To find the Meridional Difference of Latitude. 

Multiply the Distance in miles, corresponding to the Meridional Difference of 
Latitude, by the miles in a degree of longitude, in latitude equal to the Course, and 
divide the product by 60 ; the quotient will be the number of miles in the Meri- 
dional Difference of Latitude. 

M. To find the Distance.^ 

Multiply the Proper Difference of Latitude in miles, by 60, and divide the product 
by the miles in a degree of longitude, in latitude equal to the Course ; the quotient 
will be the number of miles in the Distance. 

EXAMPLE. 

A ship from the Latitude of 4° 03^ North sails South-East by East, until her 
Difference of Longitude is 900 miles ; required her Latitude in and the Distance 
sailed. (The ship must have crossed the Equator to make 900 miles Difference of 
Longitude on this Course.) 

\st. To find a Distance corresponding to the Meridional Difference of Latitude. 

By the Table of Points opposite to the Compass, on page 10, 

The Complement of the Course, viz., 3 points = 33° 45' gives 49.88810 the number of miles in a degree of 

longitude in latitude equal to it. 

900 V 60 

= 1082.40242* the number of miles in a Distance corresponding to the Meridional Dif. of Latitude 

49.88810 

2d. To find the Meridional Difference of iMtitvde. 
By the Table of Points opposite to the Compass, on page 10, 

The Course S. E. by E. 5 points = 56° 15' gives 33.33418 the number of miles in a degree of longitude in 

latitude equal to it. 

1082.40242 X 33.33418 



60 



601.34995 the number of miles in the Meridional Difference of Latitude. 



Latitude left 4° 03' N Meridional Parts 243 as per Table TIT., of the Epitome, 

601 Meridional Difference of Lat. 



5 57 8. Answer, opposite to 358 Meridional Parts, a.^ per Table III. 



Proper Difference of Latitude 10 00 
60 

Proper Dif. of Lat. in miles... 600 

Zd. To find the Distance. 

600 X 60 

— -— — — = 1079.97257 the number of miles in the Distance ^ 1080. Answer. 
33.33418 



* This case cannot be worked by Middle Latitude Sailing. Bowditch has omitted to give it ! "Why ? 

t The Distance may be found otherwise, by the Departure, thus : 

As Mer. Dif. of Lat. : Dif of Long. : : Proper Dif of Lat. : Departure. 
601.34995 : 900 : : 600 : 897.97903 



■j/(Prop. Dif of Lat.)2 + (Departure)2 = -|/(600)2 + («97 .97963)2 = |/360000 + 80(5367.4158949369 = 

l/ll66367.314S949369 = 1079.98487 = 1080 mile«. Distance. Answer. 



72 MERCATOR'S SAILING. 

QUESTIONS 

To exercise the Learner i?i the foregoing Rules. 

QUESTION I. 

A ship in the Latitude of 49° 57^ North, and the Longitude of 15° 16^ "West, sails 
South-^Veste^ly until her Departure is 789 miles, and by observation is in the Lati- 
tude of 39° 20^ North ; required her Distance, Course, and Longitude in. 

BY CASE II 

Latitude left __ 4^o 57' N Meridional Parts 3470, as per Table III., of the Epitome, 

Latitude in ^ 20 N " " 2571" " " " " " 

Proi>er Difference of Latitude 10 37 Meridional Dif. of Latitude 899 

60 

Prop. Dif. of Lat. in miles 637 

\st. To Jind the Distance. 

v/ (637)2 + (789)2 = t/405769 -f 622521 =|/l028290 = 1014.04635 the number of miles in the Distance = 
*^ V V jQj^^^ Answer. 

2d. To Jind the Course. 

= 37.69058 the number of miles in a degree of longitude in latitude equal to the Course, viz.: 



lOU.046a5 



51° 05' W., as per Example XXVI., page 25. Answer, . 



Zd. To Jind the Difference of Longitude. 

8d9 y 60 

„V-— — - = 1431.12682 the number of miles in a Distance corresponding to the Meridional Dif, of Latitude. 

37.690.dS 

By the Table of Miles in a degree of Longitude in every Latitude, as per Example XXX., page 18, 

The Complement of the Course, viz., 38° 55 gives 46.68360 the number of miles in a degree of longitude in 

latitude equal to it. 

1431.126S2 X 46.68360 1114 
—■ =1113.50253 the number of miles in the Dif. of Longitude = -— = 18° 34' W., as per Q. 

Which added to the Longitude left 15 16 W. 

GiTBs the Longitude in _. 33 50 W. 

Answer. 

QUESTION XL 

A ship in the Latitude of 42° 10^ North and the Longitude of 58° 51^ West, 
sails South-West by South 591 miles ; required the Latitude and Longitude in. 

BY CASE III. 

\st. To Jind the Proper Difference of Latitude. 

i5y the Table of Points opposite to the Compass, on page 10, 

The Course S."W. by S. 3 points = 33° 45' gives 49.88810 the number of miles in a degree of longitude in 

latitude equal to it. 

501 y 49.88810 491 

— = 491.39779 the number of miles in the Proper Dif of Lat. = — — = 8° 11' S., as per Question. 

60 60 

Latitude left 42° .30' X Meridional Parts 2822 per Table lU., of the Epitome. 

Propor Dif. of Latitude 8 11 S. 

Latitudei.ii .,,.^. 34 19 X. " " 2194 « " « « " « 

Answer. 

Meridional Dif. of Latitude 628 

Id. To find the Difference of Longitude. 

62^ X fiO 

■ , = 755.29034 the number of miles in a Distance corresponding to the Meridional Dif. of Latitude. 

By the Table of Poiats opposite to the Compass, on page 10, 

The Complement of the Course, viz. 5 points = 56° 15', gives 33.33418 the number of miles in a degree of 

longitude in latitude equal to it. 

75'i 29034 V 3.3 '53418 420 

" ' ' — , ' •' = 419.61640 the number of miles in the Dif of Long. = — - = 7° 00' W. as per Question. 

60 60 

Which added to the Longitude left 58 51 W. 

Gives the Longitude in 65 51 W. 

Answer. 



MERCATOR'S SAILING. 73 

QUESTIONS 

To exercise the Learner in the foregoing Rules. 
QUESTION III. 

A sliip from the Latitude of 49° 57^ North, and the Longitude of 30° 00^ West, 
sails S. 39° W. till she arrives in the Latitude of 45° 31^ North ; required the Dis- 
tance run and Longitude in. 

BY CASE IV. 

Latitude left 49° 57' N Meridional Parts 3470, as per Table III. of the Epitome. 

in 45 31 N " " 3074 " " « " 



Proper Dif. of Latitude 4 26 Merid. Dif. of Lat. 

60 



Prop. Dif. of Lat. in miles 266 

l5^. To find the Distance. 

By the Table of Miles in a degree of Longitiido in every Latitude, on page 9, 
The Course S. 39° "W. gives 46.62882 the number of miles in a degree of longitude in latitude equal to it. 

• = 342.27759 the number of miles in the Distance = 342.3 Answer. 

46-62882 

2d. To find the Difference of Longitude. 

^ = 509.55611 the number of miles in a Distance corresponding to the Meridional Dif. of Latitude. 

46.62S(>2 

By the Table of Miles in a degree of Longitude in every Latitude, on page 9, 

The Complement of the Course, viz., 51°, gives 37.75922 the number of miles in a degree of longitude in 

latitude equal to it. 

609.55611 X 3/ .75922 ^ ggQ g^-^Qg ^-^^ number of miles in the Dif. of Long. = ^ = 5° 21' W. as per Question. 
60 60 

Which added to the Longitude left 30 00 W. 



Gives the Longitude in 35 21 W. 

Answer. 

QUESTION IV. 

A ship from the Latitude of 50° 10^ South and the Longitude of 30° 00^ East, sails 
E.S.E. until her Departure is 957 miles; required the Distance sailed, and the Lati- 
tude and Longitude in. 

BY CASE VI. 

1st. To find the Distance. 

By the Table of Points opposite to the Compass, on page 10, 
The Complement of the Course, viz., 2 Points = 22° 30' gives 55.43268 the number of miles in a degree of 

longitude in latitude equal to it. 

957 X 60 

— — = 1035.85105 the number of miles in the Distance = 1036. Answer. 

55.43268 

2d. To find the Proper Difference of Latitude. 

By the Table of Points opposite to the Compass, on page 10, 

The Course E.S.E. 6 points = 67° 30' gives 22.96101 the number of miles in a degree of longitude in latitude 

equal to it. 

1035.85105 X 22.9610 1 ^ ggg^Qg^j^Q ^^^ number of miles in the Proper Dif. of Lat. = — = 6° 36' S., as per Qu. 
60 ^ 60 ' ^ 

Latitude left 50° 10' S Meridional Parts 3490, as per Table III. of the Epitome. 

Proper Dif. of Latitude 6 36 S. 

Latitude in 56 46 S « " 4157 

Answer. 

Meridional Dif. of Lat. 667 

2>d. To find the Difference of Longitude. 
cr,7 X 60 



22.96101 



1742.95469 the number of miles in a Distance corresponding to the Meridional Dif. of Latitude. 



1742.95409 X 55.4-3268 1610 
^ = 1610.27749 the number of miles in the Dif of Long. = — - = 26° 50' E., as per Qu. 

Which added to the Longitude left 30 00 E. 

Gives the Longitude in 56 50 E. 

Answer. 



74 MERCATOR'S SAILING. 

QUESTIONS 
7b exercise (Tie Learner in the foregoing Rules. 

QUESTION V. 

A ship in the Latitude of 49° 3(K North, and the Longitude of 25° 00^ West, sails 
South-Easterly 645 miles, making 500 miles Departure ; required the Course, and 
the Latitude and Longitude in. 

BY CASE YII. 

\st. To find the Course. 

600 X 60 

— ^jT — = 46.51163 the number of miles in a degree of longitude, in latitude equal to the Complement 

»**^ of the Course, viz., 39° 11', as per Example XXVII., page 25. Subtract the Complement 

from 90° OC, and the remainder will be the Course, S. 50° W E., as per Question. Ana. 

2d. To find the Proper Difference of Latitude. 

Bt the Table of Miles in a degree of Longitude in every Latitude, as per Example XXTI., page 17, 
The Course, S. 50° 49' E., gives 37.90817 the number of miles in a degree of longitude in latitude equal to it. 

645 X 37.90817 408 

— == 407.51283 the number of miles in the Proper Dif. of Lat. = — - = 6° 48' S., as per Question. 

DO 60 

Latitude left _ 49° BC N Meridional Parts 3428, as per Table in., of the Epitome, 

Proper Difference of Latitude 6 48 S. 

Latitude in 42 42 N " " 2839" " " " " " 

Answer. 

Meridional Dif of Latitude 5S9 

Zd. To find the Difference of Longitude. 

589 X 60 

•————- = 932.25286 the number of miles in a Distance corresponding to the Meridional Dif. of Latitude. 



— 722.67667 the number of miles in the Dif. of Long. = — - = 12° 03' E., as per Qn. 
60 60 



932.25286X46.51163 „____, , , -, ■ ,^. r,-r rr '^3 

imber of miles in the Dif. of Long. = — 
^ 60 
"Which subtracted from the Longitude left 25 00 W. 

Gives the Longitude in 12 57 W. Answer 



COMPOUND COURSES. 

Given the latitude and longitude of the place sailed from, and the different 
courses and distances sailed, to lind the difference of latitude made good, the depar- 
ture made good, the latitude in, the distance made good, the course made good, the 
difference of longitude made good, and the longitude in. 

RULE. 

Isi. To find the Difference of Latitude made good, and the Departure made good. 

Make a Traverse Table of the different Courses and Distances sailed, and by it 
find the Difference of Latitude made good, and the Departure made good, by Part II. 
of the Rule for Traverse Sailing, on page 35. 

2d. To find the Distance made good. 

Add the square of the Difference of Latitude in miles made good, to the square of 
the Departure in miles made good, and from the sum of these tTvo squares extract 
the square root, which will be the Distance in miles made good, by Part I. of the 
Kule for Case YI. Plane Sailing, on page 31. 

M. To find the Course made good. 

Multiply the Proper Difference of Latitude in miles made good, by 60, and divide 
the product by the Distance in miles made good, the quotient will be the number 
of miles in a degree of longitude, in latitude equal to the Course made good, by 
Part I. of the Rule for Case Y., Mercator's Sailing, on page 51. 

Ath. To find the Difference of longitude made good. 

Multiply the Meridional Difference of Latitude in miles, by 60, and divide the 
product by the number of miles in a degree of longitude, in latitude equal to the 
Course ; the quotient will be a Distance in miles, corresponding to the Meridional 
Difference of Latitude. Multiply this Distance in miles, by the miles in a degree 
of longitude, in latitude equal to the Complement of the Course, and divide the pro- 
duct by 60 ; the quotient will be the number of miles in the Difference of Lrm- 
gitude'made good, by Part II. of the Rule for Case Y., Mercator^s Sailing, on p. 51. 



MERCATOR'S SAILING. 



75 



EXAMPLE. 

A ship takes her departure from Cape Henlopen, in the Latitude of 38° 47^ North 
and the Longitude of 75° 05^ West, bcarino; W. by N., distant 20 miles, and then 
Bails on the foUo^ving true courses, viz., E.N.E. 15 miles, S.E. 26 miles, South 16 
miles, W.S.W. 6 miles, N.W. 10 miles, and East 30 miles ; required the Difference 
of Latitude made good, the Departure made good, the Latitude in, the Distance 
made good, the Course made good, the Difference of Longitude made good, and the 
Longitude in. 

Eemark.—See Compound Courses, in Middle Latitude Sailing, where this Ex- 
ample is worked out by that method. 

1st. To find the Difference of Latitude made good, and the Departure made good. 

Having calculated all the Differences of Latitude and Departures made on the 
several Courses and Distances sailed, by Case I., Plane Sailing, arrange them, ac- 
cording to the Rule, in the form of a 

TRAVERSE TABLE. 





i Difference of Latitude. 


1 
Departure. 


Course. 


Distance. \ N. , " S. 


E. 1 W. 


E. by S. 


20 


3.9 


19.6 j 


E.N.E. 


15 


5.7 




13.9 


S.E. 


26 




18.4 


18.4 


South 


16 




16.0 




w.s.w. 


6 




2.3 


5.5 


N.W. 


10 


7.1 




7.1 


East 


30 






30.0 






12.8 


40.6 

12.8 


81.9 i 12.6 
12.6 1 


Remain 


der 27.8 


69.3 Remainder 


i 


Difference of Latitude made good. 


Departure made good. 



2d. To find the Distance made good. 



1 / (27.8)2 -t- ita.3)2 = -I./772.84 + -tSO-'.-ta = -1/5575.33 = 74.66813 the immber of miles in the Distanco = 
*■ " * 74.7 Answer. 

2>d. To find the Course made good. 

27.S X ^0 

-— = '22.."3S85 th-e nnmber of miles in a degree of londtude in latitude equal to the Course made 

' 4.ti(.^ !3 „noA, viz., S. 68^^ OS' E.. as per Example XX VIII., p. 25. Answer. Because the Difference 

of Latitude made good is to the Southward, and the Departure made good is to the 
Eastward. 

r^atitude of Cane Henlopen 3S° 47' N Meridional Parts 252S, as per Table III., of the Epitome. 

Proper Differeiice of Latitude... 2S S. 



Latitude in. 



38 19 X. 

Answer. 



» 2492 
Merid. Dif. of Lat. 36 



Ath. To find the Difference of Longitude made good. 

90.69253 the number of miles in a Distance corresponding to the Meridional Dif. of Latitude. 



36 X 60 
22.33S85 

By the Table of Miles in a degi-ee of Longitude in every Latitude, as per Example XXXI., page 18, 
The Complement of the Courst, viz. 21*^ 52', gives 55.68313 the number of miles in a degi-ee of longitude 



()fl.K9253 X 55.68313 
GO 



in latitude equal to it. 
90 
"60 



89.73571 the number of miles in the Dif. of Longitude = -~ = 1^ 30' E., as per Ex. 
Which subtracted from the Longitude left 75 05 W. 

Gives the Longitude in „. 73 35 W. 

Answer 



l^ "i?N \ 



\ 



